# A universal compactness certificate for a finite topology [← Guide index](https://ocme.evemisslab.com/en/derivations/) · [Original MKO](https://ocme.evemisslab.com/en/mko/mko-compact-topological-space/) This is an owner-adopted natural-language teaching guide. Finite/arithmetic checks are not universal proofs; no Lean was run, and review scope is recorded per revision. Original source status labels are retained as historical data. Formulas retain exact TeX; no new semantic AST is invented. derivation-mko-compact-topological-space-01 · ocme-new142-reader-adoption-20261021 ## 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 S01 · impliesBefore \varnothing\subset\{a\}\subset\{a,b\}\subset X After \tau\text{ is a complete topology with }4\text{ members} Conditions: none Rule: topology · Uses: G1 Leaf basis: topology The finite chain includes both endpoints; nonempty unions/intersections take largest/smallest members and verify every axiom. S02 · impliesBefore X=\bigcup_{i\in I}U_i,\quad U_i\in\tau After c\in U_j\text{ for some }j\in I Conditions: none Rule: sets · Uses: G1, S01 Leaf basis: sets For an arbitrary cover, some original member must cover c. S03 · impliesBefore c\in U_j\in\tau After U_j=X Conditions: none Rule: sets · Uses: S01, S02 Leaf basis: sets Only X in the complete four-member family contains c. S04 · impliesBefore J=\{j\}\subseteq I After \bigcup_{i\in J}U_i=X,\quad |J|=1 Conditions: none Rule: sets · Uses: S03 Leaf basis: sets Retain the original index j to obtain a one-member finite subcover. S05 · definitionBefore \forall\text{ open covers }(U_i)_{i\in I},\ \exists J\subseteq I\text{ finite} After X\text{ is compact} Conditions: none Rule: definition_here · Uses: S02, S04 Leaf basis: definition_here The starting cover was arbitrary, so this establishes the definition, not merely one successful audit. S06 · impliesBefore D=\{U_i:i\in I\}\subseteq\tau After |D|\le4 Conditions: none Rule: sets · Uses: S01 Leaf basis: sets Another finite-family proof deduplicates distinct opens and selects one original index for each. S07 · impliesBefore (\{a\},\{a,b\},X) After \{X\}\text{ is a subcover of this supplied cover} Conditions: none Rule: sets · Uses: S01 Leaf basis: sets The third member of this supplied cover gives a subcover, consistent with but not replacing the universal proof. S08 · impliesBefore X=\varnothing\text{ in the empty-space boundary} After J=\varnothing\text{ covers }X Conditions: none Rule: sets · Uses: S05 Leaf basis: sets For the empty-space boundary the empty subcover satisfies the same definition. ## 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 entryStart with an arbitrary cover and choose original indices depending on it. Finite versus sampledA 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 [Same-source full JSON (formulas, ASTs, controls and revision log)](https://ocme.evemisslab.com/data/derivations/coverage-20261021/guides/derivation-mko-compact-topological-space-01.json)