A universal compactness certificate for a finite topology

Goal and scope

Prove every open cover of the three-point chain topology has a finite subcover.

A universal finite-space argument plus one cover audit; no classification of infinite spaces.

Givens and symbols

• sets
• topology
• definition_here

• G1 — X={a,b,c}, tau={empty,{a},{a,b},X}; a cover may be any indexed family (U_i) indexed by I.

• I — arbitrary cover index set
• J — finite subset of original indices

Rules and sources

• mko_definition · cited — Validate the complete finite carrier and topology, then use the universal finiteness argument: there are finitely many distinct open sets, so every open cover has a finite subcover. Supplied-cover audits are separate; success on one cover does not establish compactness of an arbitrary or infinite space. · mko-compact-topological-space
• sets · prerequisite — Compute unions, intersections, complements, inclusion and preimages element by element.
• topology · prerequisite — A topology contains empty and whole sets and is closed under arbitrary unions and finite intersections; a finite family admits exhaustive checks.
• definition_here · prerequisite — Compactness: for every indexed open cover (U_i) indexed by I, a finite J contained in I selects original members still covering the whole space.

Main derivation

\varnothing\subset\{a\}\subset\{a,b\}\subset X

\tau\text{ is a complete topology with }4\text{ members}

X=\bigcup_{i\in I}U_i,\quad U_i\in\tau

c\in U_j\text{ for some }j\in I

c\in U_j\in\tau

U_j=X

J=\{j\}\subseteq I

\bigcup_{i\in J}U_i=X,\quad |J|=1

\forall\text{ open covers }(U_i)_{i\in I},\ \exists J\subseteq I\text{ finite}

X\text{ is compact}

D=\{U_i:i\in I\}\subseteq\tau

|D|\le4

(\{a\},\{a,b\},X)

\{X\}\text{ is a subcover of this supplied cover}

X=\varnothing\text{ in the empty-space boundary}

J=\varnothing\text{ covers }X

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · S05 — This is a separate empty-carrier boundary, not a replacement of this example's X.
• S07 · premise_audit · S01 — This instance and the universal conclusion at S05 are different evidence levels.
• S06 · premise_audit · S01 — Arbitrarily repeated members do not increase the number of distinct opens.
• S05 · premise_audit · S02, S04 — Arbitrariness is the key premise and cannot be recovered from one numerical success.
• S04 · premise_audit · S03 — U_j has not been replaced by a set absent from the original family.
• S03 · premise_audit · S01, S02 — This follows from exhaustive open-set classification, not one inspected member.
• S02 · premise_audit · G1, S01 — The selected j may depend on the cover, matching the required quantifier order.
• S01 · premise_audit · G1 — Finiteness holds for the complete family, not a sample.

Error controls and negative controls

Universal entry

Start with an arbitrary cover and choose original indices depending on it.

Finite versus sampled

A complete finite-topology certificate is not a general conclusion from finite tests.

• Testing three covers proves any space compact. → reject — It does not quantify over all covers.
• One may add X to a subcover even if absent from the original cover. → reject — A subcover uses original members; here X was proved present.
• A finite sample of opens represents an entire infinite-space topology. → reject — Completeness and finiteness of the family are premises, not extrapolations.

Result and limitations

This complete finite topology is compact, with a proof for arbitrary indexed covers.

Verification and source binding

0 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: bb1992c96e3fb1b5296d4e74a562981227df547020a5dd2037c3b6c406bd4dff
• original source SHA-256: bb1992c96e3fb1b5296d4e74a562981227df547020a5dd2037c3b6c406bd4dff
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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