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Question 16712: A Set and Its Complement Have the Same Topological Boundary

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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R0032-16712 · q-20261014-advance-20261102-r04-topological-boundary-0062

Original question and givens

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

  • G001 — Use the complete finite topology=[[], [0], [0, 2], [0, 2, 4], [0, 2, 3, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Subset=[2, 4]. Give (subset,complement,boundary_a,boundary_complement,equal?).

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, 2], [0, 2, 4], [0, 2, 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

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

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

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

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

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

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

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

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

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

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

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

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

boundary 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

boundary subset:original open index=0;open=[];subset=[2, 4]

After

whole open contained in subset=Yes;running union of qualifying opens=[]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S008

S010 · implies

Before

boundary subset:original open index=1;open=[0];subset=[2, 4]

After

whole open contained in subset=No;running union of qualifying opens=[]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S009

S011 · implies

Before

boundary subset:original open index=2;open=[0, 2];subset=[2, 4]

After

whole open contained in subset=No;running union of qualifying opens=[]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S010

S012 · implies

Before

boundary subset:original open index=3;open=[0, 2, 4];subset=[2, 4]

After

whole open contained in subset=No;running union of qualifying opens=[]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S011

S013 · implies

Before

boundary subset:original open index=4;open=[0, 2, 3, 4];subset=[2, 4]

After

whole open contained in subset=No;running union of qualifying opens=[]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S012

S014 · implies

Before

boundary subset:original open index=5;open=[0, 1, 2, 3, 4];subset=[2, 4]

After

whole open contained in subset=No;running union of qualifying opens=[]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S013

S015 · implies

Before

boundary subset:ALL contained opens in original topology order=[[]]

After

topology-relative interior=[]

Conditions: none

The empty open always qualifies.The result depends on the specified topology;membership in A alone does not make a point interior.

Step identifiers and dependencies

Rule: interior_all_candidates_union_not_subset_membership · Uses: G001, S014

S016 · implies

Before

boundary subset:same-topology closure=[1, 2, 3, 4];same-topology interior=[]

After

boundary=closure minus interior=[1, 2, 3, 4]

Conditions: none

Subtract the interior from the closure,not from A itself.A point outside A can be a boundary point.No Euclidean drawing or metric is being used.

Step identifiers and dependencies

Rule: boundary_same_topology_closure_minus_interior · Uses: G001, S015

S017 · implies

Before

complement subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[0, 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, S016

S018 · implies

Before

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

After

closed set contains whole subset=No;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, S017

S019 · implies

Before

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

After

closed set contains whole subset=No;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, S018

S020 · implies

Before

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

After

closed set contains whole subset=No;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, S019

S021 · implies

Before

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

After

closed set contains whole subset=No;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, S020

S022 · implies

Before

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

After

closed set contains whole subset=No;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, S021

S023 · implies

Before

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

After

topology-relative closure=[0, 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, S022

S024 · implies

Before

complement subset:original open index=0;open=[];subset=[0, 1, 3]

After

whole open contained in subset=Yes;running union of qualifying opens=[]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S023

S025 · implies

Before

complement subset:original open index=1;open=[0];subset=[0, 1, 3]

After

whole open contained in subset=Yes;running union of qualifying opens=[0]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S024

S026 · implies

Before

complement subset:original open index=2;open=[0, 2];subset=[0, 1, 3]

After

whole open contained in subset=No;running union of qualifying opens=[0]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S025

S027 · implies

Before

complement subset:original open index=3;open=[0, 2, 4];subset=[0, 1, 3]

After

whole open contained in subset=No;running union of qualifying opens=[0]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S026

S028 · implies

Before

complement subset:original open index=4;open=[0, 2, 3, 4];subset=[0, 1, 3]

After

whole open contained in subset=No;running union of qualifying opens=[0]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S027

S029 · implies

Before

complement subset:original open index=5;open=[0, 1, 2, 3, 4];subset=[0, 1, 3]

After

whole open contained in subset=No;running union of qualifying opens=[0]

Conditions: none

The whole open set must lie in A,not merely intersect it.Unite every qualifying open to obtain the largest open subset relative to this topology.

Step identifiers and dependencies

Rule: interior_every_contained_open_original_order · Uses: G001, S028

S030 · implies

Before

complement subset:ALL contained opens in original topology order=[[], [0]]

After

topology-relative interior=[0]

Conditions: none

The empty open always qualifies.The result depends on the specified topology;membership in A alone does not make a point interior.

Step identifiers and dependencies

Rule: interior_all_candidates_union_not_subset_membership · Uses: G001, S029

S031 · implies

Before

complement subset:same-topology closure=[0, 1, 2, 3, 4];same-topology interior=[0]

After

boundary=closure minus interior=[1, 2, 3, 4]

Conditions: none

Subtract the interior from the closure,not from A itself.A point outside A can be a boundary point.No Euclidean drawing or metric is being used.

Step identifiers and dependencies

Rule: boundary_same_topology_closure_minus_interior · Uses: G001, S030

S032 · implies

Before

same carrier Fin(5);A=[2, 4];complement=[0, 1, 3];computed boundaries=[[1, 2, 3, 4], [1, 2, 3, 4]]

After

interior(A)=X minus closure(X minus A);boundary(A)=closure(A) intersect closure(X minus A);complement swaps the two factors

Conditions: none

In the same carrier,the complement of the interior is the closure of the complement.Thus the boundary is the intersection of the two closures.Complementation swaps the factors,which is the universal same-topology reason.Keep both instantiated computations and do not compare different ambient spaces.

Step identifiers and dependencies

Rule: boundary_complement_duality_same_ambient_topology · Uses: G001, S031

Reverse checks

  • S032 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S031 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S030 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S029 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S028 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S027 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S026 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S025 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S024 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S023 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S022 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S021 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S020 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S019 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S018 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S017 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S016 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S015 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S014 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S013 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S012 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S011 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S010 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • 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

([2, 4], [0, 1, 3], [1, 2, 3, 4], [1, 2, 3, 4], Yes)

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: f0232fb952d52720a1cb443d42089de8f451b8f92be0aa7eb584f4cfce4e1adf
  • candidate record SHA-256: e6c9b774f71c33047018f268d9a395c39cf822a97a27f574e5acd3b9a94f4473
  • semantic review receipt SHA-256: d64b4e08bda1c25dd393b7c7e4140a8734610a9b84ae47e6f0adcff7f30a0b8e
  • question-first witness SHA-256: 790fd6623c04ba52a6ffb0a95cea52b532a02b23bb9453926d2d9f7a825fdec8
  • Lean: NOT_RUN · human review: NOT_PERFORMED
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