Complete Rectangle-Union Topologies and Coordinate Continuity
This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Finite product topology · qf-mod48-product-topology-boundary · candidate
Canonical question JSONL · English JSONL · Validation report