# Finite Universal Compactness and Separate Supplied-Cover Audits

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-20261014-advance-20261102-r05-compact-topological-space-foundations · 2026-10-14 · 100 questions

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



## 1. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0001/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 2. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0002/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 3. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0003/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 4. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0004/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 5. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0005/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 6. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0006/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 7. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0007/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 8. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0008/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 9. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0009/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 10. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0010/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 11. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0011/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 12. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0012/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 13. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0013/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 14. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0014/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 15. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0015/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 16. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0016/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 17. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0017/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 18. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0018/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 19. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0019/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 20. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0020/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 21. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0021/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 22. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0022/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 23. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0023/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 24. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0024/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 25. Finiteness Supplies the Universal Compactness Certificate

Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).

View answer and explanation(5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology)

The complete topology has finitely many distinct open sets. Every set-family open cover is therefore finite; an indexed cover with repetitions can select representatives of its finitely many distinct members. This finite structural argument covers every open cover, unlike a single tested cover. It is not a new Lean receipt or an infinite-space compactness certificate. The exact result, retaining the requested order, is (5, 6, Yes, every_open_cover_is_a_subfamily_of_a_finite_topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0025/)

Compact topological space · qf-mod33-compact-finite-space-certificate · candidate





## 26. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0026/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 27. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0027/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 28. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0028/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 29. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0029/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 30. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0030/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 31. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0031/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 32. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0032/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 33. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0033/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 34. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0034/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 35. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0035/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 36. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0036/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 37. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0037/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 38. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0038/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 39. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0039/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 40. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0040/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 41. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0041/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 42. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0042/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 43. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0043/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 44. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0044/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 45. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0045/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 46. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0046/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 47. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0047/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 48. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0048/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 49. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0049/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 50. An Invalid Supplied Cover Does Not Refute Finite-Space Compactness

Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. Give (invalid_open_indexes,covered_union,covers_target?,valid_cover?).

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

Audit all original members and target coverage separately. A failed supplied family is not an open-cover counterexample to the universal finite-space compactness theorem. The exact result, retaining the requested order, is ([], [0, 1, 2, 3, 4], Yes, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0050/)

Compact topological space · qf-mod33-compact-cover-audit · candidate





## 51. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0051/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 52. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0052/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 53. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0053/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 54. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0054/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 55. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0055/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 56. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0056/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 57. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0057/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 58. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0058/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 59. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0059/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 60. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0060/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 61. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0061/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 62. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0062/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 63. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0063/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 64. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0064/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 65. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0065/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 66. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0066/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 67. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0067/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 68. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0068/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 69. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0069/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 70. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0070/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 71. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0071/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 72. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0072/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 73. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0073/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 74. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 2, 3, 4], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0074/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 75. A Subcover Witness Is Bound to Its Original Supplied Cover

Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,covered_union,valid_subcover?,finite_subcover_for_supplied_cover).

View answer and explanation([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover)

Read the selected original members and test target coverage. The whole original family must first qualify as an open cover; a selected good subfamily does not erase an invalid unselected original member in this protocol. The exact result, retaining the requested order, is ([1], [0, 1, 2, 3], Yes, finite_subcover_for_supplied_cover).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0075/)

Compact topological space · qf-mod33-compact-subcover-witness · candidate





## 76. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0076/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 77. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0077/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 78. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0078/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 79. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0079/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 80. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 1], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0080/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 81. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 1], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0081/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 82. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0082/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 83. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0083/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 84. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0084/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 85. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0085/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 86. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 2], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0086/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 87. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 2], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0087/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 88. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0088/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 89. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0089/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 90. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 3], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0090/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 91. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 3], [0, 2, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0091/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 92. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0092/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 93. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 3], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 3], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0093/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 94. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0094/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 95. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 4], [0, 1, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0095/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 96. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 4], original ordered cover=[[0, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0096/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 97. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 4], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0097/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 98. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 3, 4], original ordered cover=[[0, 4], [0, 1, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0098/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 99. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [0], [0, 4], [0, 3, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 2, 3, 4], original ordered cover=[[0, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0099/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate





## 100. Keep One-Cover Evidence Separate from the Finite Universal Theorem

Use the complete finite topology=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Target=[0, 1, 2, 3], original ordered cover=[[0, 1], [0, 1, 2, 3], [0, 1, 2, 3, 4]]. Every original member must be open and the original union must contain the target. A subcover selects only original family indexes, starting at 0, sorted increasingly with duplicate indexes forbidden. This supplied-cover audit is separate from the universal finite-space compactness theorem. selected=[1]. Give (selected_indexes,valid_subcover?,proves_general_compactness?,one_supplied_cover_does_not_prove_general_compactness).

View answer and explanation([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness)

The last No withholds a general compactness conclusion from this one supplied-cover audit. It does not classify this finite space as noncompact: a separate finiteness argument already supplies its universal compactness certificate. Infinite spaces cannot be classified from finitely sampled covers alone. The exact result, retaining the requested order, is ([1], Yes, No, one_supplied_cover_does_not_prove_general_compactness).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261102-r05-compact-topological-space-0100/)

Compact topological space · qf-mod33-compact-single-cover-boundary · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r05-compact-topological-space-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261102-r05-compact-topological-space-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r05-compact-topological-space-foundations/validation-report.json)