Compactness of a finite subspace

Goal and scope

Prove K={b,c} compact in its ambient space and audit a relative-open cover.

A complete finite model, without a universal closed-and-bounded substitution.

Givens and symbols

• sets
• topology
• definition_here

• G1 — X={a,b,c}, tau={empty,{a},{a,b},X}, K={b,c}.

• \tau_K — induced topology
• V_i — relative-open cover member

Rules and sources

• mko_definition · cited — A cover must cover K, and a subcover must be selected from the original cover. Passing one cover audit does not prove general compactness. For the complete finite topology and finite subset used here, finiteness supplies compactness, including the empty subset by convention. Do not extrapolate this finite enumeration to infinite spaces or identify compactness with ambient closedness and boundedness without extra hypotheses. · mko-compact-subset
• 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 — Relative opens of K are U intersect K for ambient-open U; K is compact when every relative-open cover has a finite subcover from its original indices.

Main derivation

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

\tau\text{ is a topology}

U\cap K\ (U\in\tau)

\varnothing,\varnothing,\{b\},K

\tau_K=\{\varnothing,\{b\},K\}

\tau_K\text{ is a topology on }K

K\subseteq\bigcup_{i\in I}V_i,\quad V_i\in\tau_K

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

c\in V_j\in\tau_K

V_j=K

J=\{j\}\subseteq I,\quad\bigcup_{i\in J}V_i=K

K\text{ is compact in the subspace topology}

(V_1,V_2)=(\{b\},K)

J=\{2\}\text{ covers }K;\quad\{1\}\text{ does not}

K=\varnothing\text{ as a separate boundary}

\varnothing\text{ has the empty finite subcover}

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · S03 — Coverage of the empty target is vacuous, unlike the nonempty example.
• S07 · premise_audit · G1, S03 — This instance is separate from the universal proof.
• S06 · premise_audit · S04, S05 — This does not depend on whether K is ambient open or closed.
• S05 · premise_audit · S03, S04 — The inference uses all three relative opens.
• S04 · premise_audit · S03 — j depends on the cover; the cover is not a single fixed test.
• S03 · premise_audit · S02 — The universe at this step is K.
• S02 · premise_audit · G1, S01 — Ambient witnesses need not be subsets of K.
• S01 · premise_audit · G1 — The full four-member family is checked.

Error controls and negative controls

Cover K

The relative union must contain K, not X.

Original indices

A finite subcover retains original-family members.

• {b} alone covers K. → reject — c is missing.
• Every relative open is ambient open. → reject — {b} is not in the ambient topology.
• Compact always means closed and bounded in any topology. → reject — No metric is supplied, and the general equivalence needs additional premises.

Result and limitations

K is compact in the induced topology, whose opens are empty, {b} and K.

Verification and source binding

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

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: 56c43748460b659d5e4882ad8b91e59c79a448b72613dbffe32f07c5de2c09e5
• original source SHA-256: 56c43748460b659d5e4882ad8b91e59c79a448b72613dbffe32f07c5de2c09e5
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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