# Complete Finite Topology Axioms and Open-Set Inventories

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

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

batch-20261013-advance-20261018-r01-topological-space-foundations · 2026-10-13 · 100 questions

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



## 1. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0001/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 2. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [0, 1], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0002/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 3. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 1, 2], [3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [2], [0, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0003/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 4. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0004/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 5. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0005/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 6. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [1, 2], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [0, 1, 2]], Yes, No, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0006/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 7. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [0, 1, 2], [3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [2], [1, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [0, 1, 2], [3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[2], [3], [2, 3]], none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0007/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 8. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0008/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 9. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0009/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 10. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [0, 1, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [1, 3], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0010/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 11. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0011/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 12. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0012/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 13. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0013/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 14. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [0, 1], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 3]], Yes, No, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0014/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 15. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[1], [2], [1, 2]], none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0015/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 16. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0016/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 17. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0017/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 18. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [1, 2], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0018/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 19. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [0], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0019/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 20. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0020/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 21. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], No, Yes, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0021/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 22. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [1], [1, 3], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [1, 2, 3]], Yes, No, Yes, Yes, No, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0022/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 23. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [0], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, No, Yes, No, [[0], [2], [0, 2]], none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0023/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 24. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(4, [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, Yes, Yes, Yes, Yes, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0024/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 25. Complete Finite Topology Axiom Audit

Complete finite data: X=Fin(3)={0,…,2}. Candidate open family=[[], [0], [0, 1], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, contains empty set?, contains X?, all binary unions closed?, all binary intersections closed?, topology?, union counterexample or none, intersection counterexample or none).

View answer and explanation(3, [[], [0], [0, 1], [0, 1, 2]], Yes, Yes, Yes, Yes, Yes, none, none)

Check every ordered pair, including diagonal pairs, and the empty/full requirements separately. Finiteness makes the complete family check sufficient for this finite model, not a general formal proof. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, [[], [0], [0, 1], [0, 1, 2]], Yes, Yes, Yes, Yes, Yes, none, none).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0025/)

Topological space · qf-ext18-topology-axiom-audit · candidate





## 26. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0026/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 27. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0027/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 28. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0028/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 29. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0029/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 30. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0030/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 31. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0031/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 32. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0032/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 33. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0033/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 34. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0034/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 35. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0035/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 36. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0036/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 37. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0037/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 38. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0038/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 39. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0039/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 40. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0040/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 41. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0041/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 42. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0042/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 43. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0043/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 44. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0044/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 45. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0045/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 46. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0046/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 47. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0047/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 48. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0048/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 49. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0, 1, 2, 3]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0049/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 50. Every Open, Closed, and Clopen Set in a Finite Topology

Complete finite data: X=Fin(3)={0,…,2}. Candidate open family=[[], [0], [0, 1], [0, 1, 2]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. This family is required to be an actual topology. Give (n, open family, closed family, clopen family), using complements relative to X.

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

Take all complements in the fixed underlying set and intersect the resulting closed family with the open family. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (3, [[], [0], [0, 1], [0, 1, 2]], [[], [2], [1, 2], [0, 1, 2]], [[], [0, 1, 2]]).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0050/)

Topological space · qf-ext18-topology-open-closed-inventory · candidate





## 51. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, discrete).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0051/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 52. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], Yes, indiscrete).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0052/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 53. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0053/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 54. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0054/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 55. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0055/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 56. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0056/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 57. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0057/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 58. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0058/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 59. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0059/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 60. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0060/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 61. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0061/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 62. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0062/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 63. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0063/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 64. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0064/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 65. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0065/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 66. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0066/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 67. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0067/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 68. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0068/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 69. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0069/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 70. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0070/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 71. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0071/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 72. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0072/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 73. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0073/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 74. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0074/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 75. Topology Qualification before Discrete or Indiscrete Classification

Complete finite data: X=Fin(4)={0,…,3}. Candidate open family=[[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize repeated elements/subsets as sets and order subsets by their binary masks sum(2^x). A pair counterexample is the first ordered pair in that order. The open family and underlying set are different objects; sampling is not an axiom check. Give (n, normalized family, topology?, invalid-candidate/discrete/indiscrete/other-topology). Classify discrete first; on an empty X the discrete and indiscrete families coincide.

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

Qualify the candidate before classification; the discrete family is the entire powerset, whereas the indiscrete family contains only the empty set and X. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, other-topology).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0075/)

Topological space · qf-ext18-topology-discrete-indiscrete · candidate





## 76. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0076/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 77. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0077/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 78. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0078/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 79. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0079/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 80. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0080/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 81. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0081/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 82. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [0, 1, 2], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0082/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 83. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0083/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 84. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0084/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 85. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0085/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 86. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0086/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 87. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0087/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 88. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0088/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 89. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [0, 1], [0, 1, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0089/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 90. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0090/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 91. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0091/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 92. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 2], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0092/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 93. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0093/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 94. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [1, 2], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0094/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 95. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0095/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 96. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [0], [0, 3], [0, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0096/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 97. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [1], [1, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0097/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 98. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [2], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0098/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 99. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0, 1, 2, 3]], B=[[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0, 1, 2, 3]], [[], [3], [2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0099/)

Topological space · qf-ext18-topology-finer-coarser · candidate





## 100. Compare Two Qualified Topologies on the Same Finite Set

Use X=Fin(4)={0,…,3} and complete actual topologies A=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], B=[[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]]. Normalize subsets by binary-mask order. Give (n, A, B, A⊆B?, B⊆A?, equal/A-coarser/B-coarser/incomparable). Compare families of open sets, not underlying-set sizes.

View answer and explanation(4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser)

Both candidates must satisfy the topology axioms. A⊆B means B is finer; equality and incomparability are distinct outcomes. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (4, [[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]], [[], [0], [1], [0, 1], [2], [0, 2], [1, 2], [0, 1, 2], [3], [0, 3], [1, 3], [0, 1, 3], [2, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]], Yes, No, A-coarser).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261018-r01-topological-space-0100/)

Topological space · qf-ext18-topology-finer-coarser · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261018-r01-topological-space-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261013-advance-20261018-r01-topological-space-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261018-r01-topological-space-foundations/validation-report.json)