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Question 16539: Closure Membership May Hold outside the Original Subset

On this page
  1. Original question and givens
  2. Definition and scope
  3. Step-by-step derivation
  4. Reverse checks
  5. Result
  6. This question review and source binding

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R0032-16539 · q-20261014-advance-20261101-r02-closure-and-interior-0064

Original question and givens

Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Every subset is an increasing duplicate-free carrier list. First check the unique empty/whole opens and all union/intersection closure obligations; finite topology suffices for exhaustive closure checks. Complements refer to this carrier. No metric is silently supplied. Subset=[1, 3], point=4. Give (point,closure,in_closure?,topology_relative_closure).

  • G001 — Use the complete finite topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Every subset is an increasing duplicate-free carrier list. First check the unique empty/whole opens and all union/intersection closure obligations; finite topology suffices for exhaustive closure checks. Complements refer to this carrier. No metric is silently supplied. Subset=[1, 3], point=4. Give (point,closure,in_closure?,topology_relative_closure).

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

Step-by-step derivation

S001 · implies

Before

same complete carrier Fin(5);original ordered complete topology=[[], [0], [0, 3], [0, 1, 3], [0, 1, 3, 4], [0, 1, 2, 3, 4]]

After

unique empty/whole opens=Yes;ordered open pairs checked=36;first union failure=null;first intersection failure=null

Conditions: none

Each subset is increasing and duplicate-free.The topology has unique empty/whole opens and closure under every binary union/intersection.Finitely many distinct opens reduce arbitrary unions to finite ones,handled by binary closure and the empty set.No metric is supplied.Keep original topology order.

Step identifiers and dependencies

Rule: topology_complete_finite_parent_all_pairs · Uses: G001

S002 · implies

Before

selected subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[1, 3]

After

closed set contains whole subset=Yes;running intersection of qualifying closed sets=[0, 1, 2, 3, 4]

Conditions: none

Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection.

Step identifiers and dependencies

Rule: closure_every_closed_superset_original_open_order · Uses: G001, S001

S003 · implies

Before

selected subset:original open index=1;open=[0];same-carrier closed complement=[1, 2, 3, 4];subset=[1, 3]

After

closed set contains whole subset=Yes;running intersection of qualifying closed sets=[1, 2, 3, 4]

Conditions: none

Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection.

Step identifiers and dependencies

Rule: closure_every_closed_superset_original_open_order · Uses: G001, S002

S004 · implies

Before

selected subset:original open index=2;open=[0, 3];same-carrier closed complement=[1, 2, 4];subset=[1, 3]

After

closed set contains whole subset=No;running intersection of qualifying closed sets=[1, 2, 3, 4]

Conditions: none

Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection.

Step identifiers and dependencies

Rule: closure_every_closed_superset_original_open_order · Uses: G001, S003

S005 · implies

Before

selected subset:original open index=3;open=[0, 1, 3];same-carrier closed complement=[2, 4];subset=[1, 3]

After

closed set contains whole subset=No;running intersection of qualifying closed sets=[1, 2, 3, 4]

Conditions: none

Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection.

Step identifiers and dependencies

Rule: closure_every_closed_superset_original_open_order · Uses: G001, S004

S006 · implies

Before

selected subset:original open index=4;open=[0, 1, 3, 4];same-carrier closed complement=[2];subset=[1, 3]

After

closed set contains whole subset=No;running intersection of qualifying closed sets=[1, 2, 3, 4]

Conditions: none

Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection.

Step identifiers and dependencies

Rule: closure_every_closed_superset_original_open_order · Uses: G001, S005

S007 · implies

Before

selected subset:original open index=5;open=[0, 1, 2, 3, 4];same-carrier closed complement=[];subset=[1, 3]

After

closed set contains whole subset=No;running intersection of qualifying closed sets=[1, 2, 3, 4]

Conditions: none

Complement every original open in the same carrier to obtain all closed sets.Keep exactly those containing all of A,and intersect every qualifying set rather than one convenient superset.A point outside A can still lie in the final intersection.

Step identifiers and dependencies

Rule: closure_every_closed_superset_original_open_order · Uses: G001, S006

S008 · implies

Before

selected subset:ALL qualifying closed supersets in original complement order=[[0, 1, 2, 3, 4], [1, 2, 3, 4]]

After

topology-relative closure=[1, 2, 3, 4]

Conditions: none

The whole carrier is a closed superset,so the family is present.The intersection is the least closed superset in this topology.Original order affects the candidate list,not its set value.

Step identifiers and dependencies

Rule: closure_all_candidates_intersection_not_metric_guess · Uses: G001, S007

S009 · implies

Before

original point=4;original subset=[1, 3];complete computed set=[1, 2, 3, 4]

After

membership=Yes;scope=topology_relative_closure

Conditions: none

Compute the entire closure/interior before testing the original point.Do not substitute original-subset membership or an unspecified distance.

Step identifiers and dependencies

Rule: topology_point_test_computed_closure_or_interior · Uses: G001, S008

Reverse checks

  • S009 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S008 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S007 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S006 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S005 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S004 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

(4, [1, 2, 3, 4], Yes, topology_relative_closure)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.

Source hashes and review receipts
  • input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
  • original question SHA-256: 439ca65ac68485e78bd3a87a4ae0c052915193882a511feec4729018f3b4dd8d
  • candidate record SHA-256: ba872441d27d37c759fd64967674fca756cdd4a75eae32f2e6cd6428164d9aae
  • semantic review receipt SHA-256: d64b4e08bda1c25dd393b7c7e4140a8734610a9b84ae47e6f0adcff7f30a0b8e
  • question-first witness SHA-256: 790fd6623c04ba52a6ffb0a95cea52b532a02b23bb9453926d2d9f7a825fdec8
  • Lean: NOT_RUN · human review: NOT_PERFORMED
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