Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).
G001 — Use the complete finite topology=[[], [0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]] on the same carrier Fin(5)={0,…,4}. Each subset is an increasing duplicate-free carrier list. Validate the unique empty/whole opens and all finite union/intersection closure obligations first. All complements, closed sets, closures, interiors and boundaries refer to this same carrier/topology; no metric or visual boundary is silently supplied. Give (carrier_size,open_set_count,compact?,every_open_cover_is_a_subfamily_of_a_finite_topology).
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.
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.
complete finite topology contains exactly6 distinct opens on Fin(5)
After
for ANY set-family open cover U,each member lies in this finite topology;U is finite and covers by itself
Conditions: none
Quantify over any open cover,not a test suite.Every member belongs to the supplied finite topology,so the set of distinct members is finite and already constitutes a finite subcover.
same finite topology with6 distinct opens;an arbitrary INDEXED cover may repeat opens
After
select one original index per distinct open used;only finitely many choices;the representative family retains the same union
Conditions: none
An indexed cover may repeat opens infinitely,but only finitely many distinct opens occur.One original index per distinct open retains the union and gives a finite subcover.Do not confuse infinitely many indices with infinitely many distinct opens or extend this argument to arbitrary infinite spaces.
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.