# Complete Rectangle-Union Topologies and Coordinate Continuity

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261015-advance-20261117-r04-finite-product-topology-foundations · 2026-10-15 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], []], 2, 2, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], []], 2, 2, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0001/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 2. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0002/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 3. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0003/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 4. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0004/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 5. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0005/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 6. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0006/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 7. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [0], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0007/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 8. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0008/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 9. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0009/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 10. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0010/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 11. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0011/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 12. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [0], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], [[2, 100], [2, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], [[2, 100], [2, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0012/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 13. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0013/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 14. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0014/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 15. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0015/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 16. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [1], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], [[2, 100], [2, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], [[2, 100], [2, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0016/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 17. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], []], 3, 3, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0017/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 18. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0018/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 19. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0019/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 20. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0020/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 21. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0021/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 22. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[1, 100], [1, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0022/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 23. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [0], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], 6, 6, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[1, 100], [1, 101]], []], 6, 6, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0023/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 24. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[2, 100], [2, 101]], []], 4, 4, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0024/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 25. The Entire Product Topology Contains Every Union of Basic Rectangles

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (product_points, basic_rectangles, product_topology, rectangle_count, open_count, topology_certificate, certificate).

View answer and explanation([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[2, 100], [2, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions)

Retain the complete typed Cartesian carrier, all distinct rectangles and the complete union-generated topology in source canonical order. Rectangles alone need not be closed under unions or exhaust all opens. This exact family uses two nonempty factors and no zero-factor convention. The exact result, retaining the requested order, is ([[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[2, 100], [2, 101]], []], [[[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], [[0, 100], [0, 101], [1, 100], [1, 101]], [[2, 100], [2, 101]], []], 5, 5, Yes, rectangles_generate_opens_by_arbitrary_unions).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0025/)

Finite product topology · qf-mod48-product-rectangle-generation · candidate





## 26. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0026/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 27. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0027/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 28. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0028/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 29. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0029/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 30. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0030/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 31. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0031/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 32. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [0], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0032/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 33. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0033/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 34. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0034/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 35. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0035/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 36. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0036/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 37. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [0], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 2], [[0, 100], [0, 101], [2, 100], [2, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0037/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 38. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0038/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 39. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0039/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 40. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0040/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 41. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [1], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[1, 2], [[1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0041/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 42. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0042/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 43. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0043/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 44. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0044/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 45. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0045/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 46. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0046/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 47. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0047/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 48. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [0], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[0], [[0, 100], [0, 101]], Yes], [[1], [[1, 100], [1, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0048/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 49. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0049/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 50. Each Coordinate Projection Has Open Preimages for Every Factor Open

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Give (first_projection_preimages, second_projection_preimages, first_projection_continuous, second_projection_continuous, projection_certificate, certificate).

View answer and explanation([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open)

For every open of each factor retain its entire cylinder preimage and open flag in the full generated product. Check both projections, not just their values on a sample of points. No general universal-property proof is supplied by finite enumeration. The exact result, retaining the requested order, is ([[[], [], Yes], [[0, 1, 2, 3], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes], [[0, 1, 2], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101]], Yes], [[0, 1], [[0, 100], [0, 101], [1, 100], [1, 101]], Yes], [[2], [[2, 100], [2, 101]], Yes]], [[[], [], Yes], [[100, 101], [[0, 100], [0, 101], [1, 100], [1, 101], [2, 100], [2, 101], [3, 100], [3, 101]], Yes]], Yes, Yes, Yes, all_projection_preimages_are_open).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0050/)

Finite product topology · qf-mod48-product-projection-continuity · candidate





## 51. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0051/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 52. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0052/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 53. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0053/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 54. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0054/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 55. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0055/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 56. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0056/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 57. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1], [0], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0057/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 58. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0058/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 59. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0059/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 60. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0060/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 61. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0061/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 62. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 2], [0], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0062/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 63. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0063/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 64. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0064/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 65. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0065/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 66. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [1, 2], [1], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0066/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 67. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0067/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 68. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0068/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 69. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0069/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 70. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0070/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 71. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [0]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0071/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 72. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0072/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 73. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [0], [1]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (No, Yes, No, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0073/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 74. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0], [1], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0074/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 75. Continuity into a Product Is Equivalent to Continuity of Both Coordinates

Use complete first-factor points=[0, 1, 2, 3], topology=[[], [0, 1, 2, 3], [0, 1, 2], [0, 1], [2]], second-factor points=[100, 101], topology=[[], [100, 101]]. Ordered product pairs retain coordinate types; qualify both entire finite topologies before generating every rectangle union. Source Z=[0, 1], topology=[[], [0, 1]], aligned coordinate maps=[0, 1] and [100, 101]. Give (first_coordinate_continuous, second_coordinate_continuous, paired_map_continuous, criterion_equivalence, paired_map_rows, certificate).

Coordinate and paired-map flags can be No while criterion_equivalence is Yes: it compares the two propositions, not universal continuity of every candidate.

View answer and explanation(Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous)

Audit each coordinate map and the paired map on the entire finite source. Compare the paired-map flag with the conjunction of coordinate flags; the equivalence can be Yes even when some continuity flags are No. Preserve map associations when sorting source labels. The exact result, retaining the requested order, is (Yes, Yes, Yes, Yes, [[0, [0, 100]], [1, [1, 101]]], map_into_finite_product_continuous_iff_coordinates_continuous).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0075/)

Finite product topology · qf-mod48-product-coordinate-criterion · candidate





## 76. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=nonrectangle_open with the complete declared model={"x_points":[0,1],"x_topology":[[],[0],[1],[0,1]],"y_points":[100,101],"y_topology":[[],[100],[101],[100,101]],"candidate_open":[[0,100],[1,101]]}. Give (mode, product_points, candidate_open, open, basic_rectangle, basic_rectangles, certificate).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(nonrectangle_open, [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [1, 101]], Yes, No, [[[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[0, 100], [1, 100]], [[0, 100]], [[0, 101], [1, 101]], [[0, 101]], [[1, 100], [1, 101]], [[1, 100]], [[1, 101]], []], open_union_of_rectangles_need_not_be_one_rectangle)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (nonrectangle_open, [[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [1, 101]], Yes, No, [[[0, 100], [0, 101], [1, 100], [1, 101]], [[0, 100], [0, 101]], [[0, 100], [1, 100]], [[0, 100]], [[0, 101], [1, 101]], [[0, 101]], [[1, 100], [1, 101]], [[1, 100]], [[1, 101]], []], open_union_of_rectangles_need_not_be_one_rectangle).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0076/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 77. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=invalid_factor with the complete declared model={"x_points":[0,1],"x_topology":[[]],"y_points":[100,101],"y_topology":[[],[100,101]],"invalid_side":"x"}. Give (mode, invalid_side, first_factor_valid, first_factor_error_or_null, second_factor_valid, second_factor_error_or_null, product_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0077/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 78. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=finite_box_no_extrapolation with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[1]],"y_points":[100,101],"y_topology":[[],[100,101]]}. Give (mode, product_point_count, rectangle_count, open_count, finite_model_certificate, finite_box_statement, infinite_extrapolation_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(finite_box_no_extrapolation, 8, 3, 3, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (finite_box_no_extrapolation, 8, 3, 3, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0078/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 79. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=ordered_pair_direction with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,1]],"y_points":[100,101],"y_topology":[[],[100,101]],"pair":[0,101]}. Give (mode, pair, reversed_pair, pair_identity_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0079/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 80. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=nonrectangle_open with the complete declared model={"x_points":[40,41],"x_topology":[[],[40],[41],[40,41]],"y_points":[140,141],"y_topology":[[],[140],[141],[140,141]],"candidate_open":[[40,140],[41,141]]}. Give (mode, product_points, candidate_open, open, basic_rectangle, basic_rectangles, certificate).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(nonrectangle_open, [[40, 140], [40, 141], [41, 140], [41, 141]], [[40, 140], [41, 141]], Yes, No, [[[40, 140], [40, 141], [41, 140], [41, 141]], [[40, 140], [40, 141]], [[40, 140], [41, 140]], [[40, 140]], [[40, 141], [41, 141]], [[40, 141]], [[41, 140], [41, 141]], [[41, 140]], [[41, 141]], []], open_union_of_rectangles_need_not_be_one_rectangle)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (nonrectangle_open, [[40, 140], [40, 141], [41, 140], [41, 141]], [[40, 140], [41, 141]], Yes, No, [[[40, 140], [40, 141], [41, 140], [41, 141]], [[40, 140], [40, 141]], [[40, 140], [41, 140]], [[40, 140]], [[40, 141], [41, 141]], [[40, 141]], [[41, 140], [41, 141]], [[41, 140]], [[41, 141]], []], open_union_of_rectangles_need_not_be_one_rectangle).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0080/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 81. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=invalid_factor with the complete declared model={"x_points":[0,1,2],"x_topology":[[]],"y_points":[100,101],"y_topology":[[],[100,101]],"invalid_side":"x"}. Give (mode, invalid_side, first_factor_valid, first_factor_error_or_null, second_factor_valid, second_factor_error_or_null, product_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0081/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 82. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=finite_box_no_extrapolation with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,1],[0],[1]],"y_points":[100,101],"y_topology":[[],[100,101]]}. Give (mode, product_point_count, rectangle_count, open_count, finite_model_certificate, finite_box_statement, infinite_extrapolation_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(finite_box_no_extrapolation, 8, 5, 5, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (finite_box_no_extrapolation, 8, 5, 5, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0082/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 83. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=ordered_pair_direction with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[2]],"y_points":[100,101],"y_topology":[[],[100,101]],"pair":[0,101]}. Give (mode, pair, reversed_pair, pair_identity_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0083/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 84. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=nonrectangle_open with the complete declared model={"x_points":[80,81],"x_topology":[[],[80],[81],[80,81]],"y_points":[180,181],"y_topology":[[],[180],[181],[180,181]],"candidate_open":[[80,180],[81,181]]}. Give (mode, product_points, candidate_open, open, basic_rectangle, basic_rectangles, certificate).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(nonrectangle_open, [[80, 180], [80, 181], [81, 180], [81, 181]], [[80, 180], [81, 181]], Yes, No, [[[80, 180], [80, 181], [81, 180], [81, 181]], [[80, 180], [80, 181]], [[80, 180], [81, 180]], [[80, 180]], [[80, 181], [81, 181]], [[80, 181]], [[81, 180], [81, 181]], [[81, 180]], [[81, 181]], []], open_union_of_rectangles_need_not_be_one_rectangle)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (nonrectangle_open, [[80, 180], [80, 181], [81, 180], [81, 181]], [[80, 180], [81, 181]], Yes, No, [[[80, 180], [80, 181], [81, 180], [81, 181]], [[80, 180], [80, 181]], [[80, 180], [81, 180]], [[80, 180]], [[80, 181], [81, 181]], [[80, 181]], [[81, 180], [81, 181]], [[81, 180]], [[81, 181]], []], open_union_of_rectangles_need_not_be_one_rectangle).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0084/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 85. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=invalid_factor with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[]],"y_points":[100,101],"y_topology":[[],[100,101]],"invalid_side":"x"}. Give (mode, invalid_side, first_factor_valid, first_factor_error_or_null, second_factor_valid, second_factor_error_or_null, product_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0085/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 86. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=finite_box_no_extrapolation with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,2],[2]],"y_points":[100,101],"y_topology":[[],[100,101]]}. Give (mode, product_point_count, rectangle_count, open_count, finite_model_certificate, finite_box_statement, infinite_extrapolation_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(finite_box_no_extrapolation, 8, 4, 4, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (finite_box_no_extrapolation, 8, 4, 4, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0086/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 87. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=ordered_pair_direction with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,2],[0],[2]],"y_points":[100,101],"y_topology":[[],[100,101]],"pair":[0,101]}. Give (mode, pair, reversed_pair, pair_identity_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0087/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 88. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=nonrectangle_open with the complete declared model={"x_points":[120,121],"x_topology":[[],[120],[121],[120,121]],"y_points":[220,221],"y_topology":[[],[220],[221],[220,221]],"candidate_open":[[120,220],[121,221]]}. Give (mode, product_points, candidate_open, open, basic_rectangle, basic_rectangles, certificate).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(nonrectangle_open, [[120, 220], [120, 221], [121, 220], [121, 221]], [[120, 220], [121, 221]], Yes, No, [[[120, 220], [120, 221], [121, 220], [121, 221]], [[120, 220], [120, 221]], [[120, 220], [121, 220]], [[120, 220]], [[120, 221], [121, 221]], [[120, 221]], [[121, 220], [121, 221]], [[121, 220]], [[121, 221]], []], open_union_of_rectangles_need_not_be_one_rectangle)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (nonrectangle_open, [[120, 220], [120, 221], [121, 220], [121, 221]], [[120, 220], [121, 221]], Yes, No, [[[120, 220], [120, 221], [121, 220], [121, 221]], [[120, 220], [120, 221]], [[120, 220], [121, 220]], [[120, 220]], [[120, 221], [121, 221]], [[120, 221]], [[121, 220], [121, 221]], [[121, 220]], [[121, 221]], []], open_union_of_rectangles_need_not_be_one_rectangle).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0088/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 89. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=invalid_factor with the complete declared model={"x_points":[0,1,2,3,4],"x_topology":[[]],"y_points":[100,101],"y_topology":[[],[100,101]],"invalid_side":"x"}. Give (mode, invalid_side, first_factor_valid, first_factor_error_or_null, second_factor_valid, second_factor_error_or_null, product_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0089/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 90. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=finite_box_no_extrapolation with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[1,2],[2]],"y_points":[100,101],"y_topology":[[],[100,101]]}. Give (mode, product_point_count, rectangle_count, open_count, finite_model_certificate, finite_box_statement, infinite_extrapolation_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(finite_box_no_extrapolation, 8, 4, 4, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (finite_box_no_extrapolation, 8, 4, 4, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0090/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 91. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=ordered_pair_direction with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[1,2],[1],[2]],"y_points":[100,101],"y_topology":[[],[100,101]],"pair":[0,101]}. Give (mode, pair, reversed_pair, pair_identity_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0091/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 92. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=nonrectangle_open with the complete declared model={"x_points":[160,161],"x_topology":[[],[160],[161],[160,161]],"y_points":[260,261],"y_topology":[[],[260],[261],[260,261]],"candidate_open":[[160,260],[161,261]]}. Give (mode, product_points, candidate_open, open, basic_rectangle, basic_rectangles, certificate).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(nonrectangle_open, [[160, 260], [160, 261], [161, 260], [161, 261]], [[160, 260], [161, 261]], Yes, No, [[[160, 260], [160, 261], [161, 260], [161, 261]], [[160, 260], [160, 261]], [[160, 260], [161, 260]], [[160, 260]], [[160, 261], [161, 261]], [[160, 261]], [[161, 260], [161, 261]], [[161, 260]], [[161, 261]], []], open_union_of_rectangles_need_not_be_one_rectangle)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (nonrectangle_open, [[160, 260], [160, 261], [161, 260], [161, 261]], [[160, 260], [161, 261]], Yes, No, [[[160, 260], [160, 261], [161, 260], [161, 261]], [[160, 260], [160, 261]], [[160, 260], [161, 260]], [[160, 260]], [[160, 261], [161, 261]], [[160, 261]], [[161, 260], [161, 261]], [[161, 260]], [[161, 261]], []], open_union_of_rectangles_need_not_be_one_rectangle).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0092/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 93. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=invalid_factor with the complete declared model={"x_points":[0,1,2,3,4,5],"x_topology":[[]],"y_points":[100,101],"y_topology":[[],[100,101]],"invalid_side":"x"}. Give (mode, invalid_side, first_factor_valid, first_factor_error_or_null, second_factor_valid, second_factor_error_or_null, product_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0093/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 94. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=finite_box_no_extrapolation with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,1,2],[1]],"y_points":[100,101],"y_topology":[[],[100,101]]}. Give (mode, product_point_count, rectangle_count, open_count, finite_model_certificate, finite_box_statement, infinite_extrapolation_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(finite_box_no_extrapolation, 8, 4, 4, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (finite_box_no_extrapolation, 8, 4, 4, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0094/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 95. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=ordered_pair_direction with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,1,2],[0,1]],"y_points":[100,101],"y_topology":[[],[100,101]],"pair":[0,101]}. Give (mode, pair, reversed_pair, pair_identity_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0095/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 96. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=nonrectangle_open with the complete declared model={"x_points":[200,201],"x_topology":[[],[200],[201],[200,201]],"y_points":[300,301],"y_topology":[[],[300],[301],[300,301]],"candidate_open":[[200,300],[201,301]]}. Give (mode, product_points, candidate_open, open, basic_rectangle, basic_rectangles, certificate).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(nonrectangle_open, [[200, 300], [200, 301], [201, 300], [201, 301]], [[200, 300], [201, 301]], Yes, No, [[[200, 300], [200, 301], [201, 300], [201, 301]], [[200, 300], [200, 301]], [[200, 300], [201, 300]], [[200, 300]], [[200, 301], [201, 301]], [[200, 301]], [[201, 300], [201, 301]], [[201, 300]], [[201, 301]], []], open_union_of_rectangles_need_not_be_one_rectangle)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (nonrectangle_open, [[200, 300], [200, 301], [201, 300], [201, 301]], [[200, 300], [201, 301]], Yes, No, [[[200, 300], [200, 301], [201, 300], [201, 301]], [[200, 300], [200, 301]], [[200, 300], [201, 300]], [[200, 300]], [[200, 301], [201, 301]], [[200, 301]], [[201, 300], [201, 301]], [[201, 300]], [[201, 301]], []], open_union_of_rectangles_need_not_be_one_rectangle).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0096/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 97. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=invalid_factor with the complete declared model={"x_points":[0,1,2,3,4,5,6],"x_topology":[[]],"y_points":[100,101],"y_topology":[[],[100,101]],"invalid_side":"x"}. Give (mode, invalid_side, first_factor_valid, first_factor_error_or_null, second_factor_valid, second_factor_error_or_null, product_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (invalid_factor, x, No, empty and full opens required, Yes, null, No, product_topology_requires_valid_factor_topologies).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0097/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 98. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=finite_box_no_extrapolation with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,1,2],[0,1],[0],[1]],"y_points":[100,101],"y_topology":[[],[100,101]]}. Give (mode, product_point_count, rectangle_count, open_count, finite_model_certificate, finite_box_statement, infinite_extrapolation_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(finite_box_no_extrapolation, 8, 6, 6, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (finite_box_no_extrapolation, 8, 6, 6, Yes, finite_product_and_box_topologies_coincide, No, no_inference_to_infinite_products).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0098/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 99. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=ordered_pair_direction with the complete declared model={"x_points":[0,1,2,3],"x_topology":[[],[0,1,2,3],[0,1,2],[2]],"y_points":[100,101],"y_topology":[[],[100,101]],"pair":[0,101]}. Give (mode, pair, reversed_pair, pair_identity_claim, boundary).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (ordered_pair_direction, [0, 101], [101, 0], No, ordered_pair_coordinates_are_typed_and_directional).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0099/)

Finite product topology · qf-mod48-product-topology-boundary · candidate





## 100. Nonrectangle Opens, Invalid Factors and Pair Direction Have Different Typed Outputs

Use mode=nonrectangle_open with the complete declared model={"x_points":[240,241],"x_topology":[[],[240],[241],[240,241]],"y_points":[340,341],"y_topology":[[],[340],[341],[340,341]],"candidate_open":[[240,340],[241,341]]}. Give (mode, product_points, candidate_open, open, basic_rectangle, basic_rectangles, certificate).

The four modes have 7/8/8/5 fields. A nonrectangle open is a genuine union, not one rectangle. Invalid factors retain each side error/null separately. Finite box equality is not extrapolated to infinite products.

View answer and explanation(nonrectangle_open, [[240, 340], [240, 341], [241, 340], [241, 341]], [[240, 340], [241, 341]], Yes, No, [[[240, 340], [240, 341], [241, 340], [241, 341]], [[240, 340], [240, 341]], [[240, 340], [241, 340]], [[240, 340]], [[240, 341], [241, 341]], [[240, 341]], [[241, 340], [241, 341]], [[241, 340]], [[241, 341]], []], open_union_of_rectangles_need_not_be_one_rectangle)

A nonrectangle witness must really be open but not a single rectangle. Invalid-factor mode retains each factor validity/error separately. Finite product and box topologies coincide here, without extrapolation to infinite products. A typed ordered pair is not identified with its reversal, which may not lie in the same product carrier. The exact result, retaining the requested order, is (nonrectangle_open, [[240, 340], [240, 341], [241, 340], [241, 341]], [[240, 340], [241, 341]], Yes, No, [[[240, 340], [240, 341], [241, 340], [241, 341]], [[240, 340], [240, 341]], [[240, 340], [241, 340]], [[240, 340]], [[240, 341], [241, 341]], [[240, 341]], [[241, 340], [241, 341]], [[241, 340]], [[241, 341]], []], open_union_of_rectangles_need_not_be_one_rectangle).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261117-r04-finite-product-topology-0100/)

Finite product topology · qf-mod48-product-topology-boundary · candidate



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