Finite Universal Compactness and Separate Supplied-Cover Audits
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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Compact topological space · qf-mod33-compact-single-cover-boundary · candidate
Canonical question JSONL · English JSONL · Validation report