# Question 16705: A Set and Its Complement Have the Same Topological Boundary [Return to catalogue page 84](https://ocme.evemisslab.com/en/questions/page/84/) [← Question derivation index](https://ocme.evemisslab.com/en/questions/) · [Original question batch](https://ocme.evemisslab.com/en/practice/batch-20261014-advance-20261102-r04-topological-boundary-foundations/#q-20261014-advance-20261102-r04-topological-boundary-0055) This teaching derivation has second-context AI mathematical and bilingual semantic review, limited to the model and assumptions stated in this question. It has no human review or new Lean proof and does not promote the original question or MKO to Canonical truth. R0032-16705 · q-20261014-advance-20261102-r04-topological-boundary-0055 ## Original question and givens Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Subset=[1, 4]. Give (subset,complement,boundary_a,boundary_complement,equal?). - G001 — Use the complete finite topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 4], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Subset=[1, 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. - [mko-topological-boundary](https://ocme.evemisslab.com/en/mko/mko-topological-boundary/) ## Step-by-step derivation S001 · impliesBefore same complete carrier Fin(5);original ordered complete topology=[[], [0], [0, 1], [0, 1, 4], [0, 1, 2, 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 dependenciesRule: topology_complete_finite_parent_all_pairs · Uses: G001 S002 · impliesBefore boundary subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[1, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S001 S003 · impliesBefore boundary subset:original open index=1;open=[0];same-carrier closed complement=[1, 2, 3, 4];subset=[1, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S002 S004 · impliesBefore boundary subset:original open index=2;open=[0, 1];same-carrier closed complement=[2, 3, 4];subset=[1, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S003 S005 · impliesBefore boundary subset:original open index=3;open=[0, 1, 4];same-carrier closed complement=[2, 3];subset=[1, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S004 S006 · impliesBefore boundary subset:original open index=4;open=[0, 1, 2, 4];same-carrier closed complement=[3];subset=[1, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S005 S007 · impliesBefore boundary subset:original open index=5;open=[0, 1, 2, 3, 4];same-carrier closed complement=[];subset=[1, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S006 S008 · impliesBefore 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 dependenciesRule: closure_all_candidates_intersection_not_metric_guess · Uses: G001, S007 S009 · impliesBefore boundary subset:original open index=0;open=[];subset=[1, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S008 S010 · impliesBefore boundary subset:original open index=1;open=[0];subset=[1, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S009 S011 · impliesBefore boundary subset:original open index=2;open=[0, 1];subset=[1, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S010 S012 · impliesBefore boundary subset:original open index=3;open=[0, 1, 4];subset=[1, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S011 S013 · impliesBefore boundary subset:original open index=4;open=[0, 1, 2, 4];subset=[1, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S012 S014 · impliesBefore boundary subset:original open index=5;open=[0, 1, 2, 3, 4];subset=[1, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S013 S015 · impliesBefore 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 dependenciesRule: interior_all_candidates_union_not_subset_membership · Uses: G001, S014 S016 · impliesBefore 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 dependenciesRule: boundary_same_topology_closure_minus_interior · Uses: G001, S015 S017 · impliesBefore complement subset:original open index=0;open=[];same-carrier closed complement=[0, 1, 2, 3, 4];subset=[0, 2, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S016 S018 · impliesBefore complement subset:original open index=1;open=[0];same-carrier closed complement=[1, 2, 3, 4];subset=[0, 2, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S017 S019 · impliesBefore complement subset:original open index=2;open=[0, 1];same-carrier closed complement=[2, 3, 4];subset=[0, 2, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S018 S020 · impliesBefore complement subset:original open index=3;open=[0, 1, 4];same-carrier closed complement=[2, 3];subset=[0, 2, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S019 S021 · impliesBefore complement subset:original open index=4;open=[0, 1, 2, 4];same-carrier closed complement=[3];subset=[0, 2, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S020 S022 · impliesBefore complement subset:original open index=5;open=[0, 1, 2, 3, 4];same-carrier closed complement=[];subset=[0, 2, 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 dependenciesRule: closure_every_closed_superset_original_open_order · Uses: G001, S021 S023 · impliesBefore 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 dependenciesRule: closure_all_candidates_intersection_not_metric_guess · Uses: G001, S022 S024 · impliesBefore complement subset:original open index=0;open=[];subset=[0, 2, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S023 S025 · impliesBefore complement subset:original open index=1;open=[0];subset=[0, 2, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S024 S026 · impliesBefore complement subset:original open index=2;open=[0, 1];subset=[0, 2, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S025 S027 · impliesBefore complement subset:original open index=3;open=[0, 1, 4];subset=[0, 2, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S026 S028 · impliesBefore complement subset:original open index=4;open=[0, 1, 2, 4];subset=[0, 2, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S027 S029 · impliesBefore complement subset:original open index=5;open=[0, 1, 2, 3, 4];subset=[0, 2, 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 dependenciesRule: interior_every_contained_open_original_order · Uses: G001, S028 S030 · impliesBefore 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 dependenciesRule: interior_all_candidates_union_not_subset_membership · Uses: G001, S029 S031 · impliesBefore 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 dependenciesRule: boundary_same_topology_closure_minus_interior · Uses: G001, S030 S032 · impliesBefore same carrier Fin(5);A=[1, 4];complement=[0, 2, 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 dependenciesRule: 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 ([1, 4], [0, 2, 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: 846088f1845665d09ca108df15471afca446787871e86caaff18106e4f767dc6 - candidate record SHA-256: 84f86b58dc8d494d61ac136fd59d232ea02606f66d3e9f795c7199d61d2037cb - semantic review receipt SHA-256: d64b4e08bda1c25dd393b7c7e4140a8734610a9b84ae47e6f0adcff7f30a0b8e - question-first witness SHA-256: 790fd6623c04ba52a6ffb0a95cea52b532a02b23bb9453926d2d9f7a825fdec8 - Lean: NOT_RUN · human review: NOT_PERFORMED [Same-source full JSON](https://ocme.evemisslab.com/data/question-derivations/questions/q-20261014-advance-20261102-r04-topological-boundary-0055.json)