S01 · case_split
Before
K=\varnothing
After
X\setminus K=X\text{ is open}Conditions: none
Rule: sets · Uses: G1
Leaf basis: sets
Handle empty K first: its complement is the open whole space.
derivation-mko-compact-subset-closed-in-hausdorff-01 · ocme-new142-reader-adoption-20261021
Prove a compact subset K of a Hausdorff space is closed.
A general cover proof, not replaced by finite instances and not proving a converse.
setstopologyhausdorff_defcompact_defG1 — X is Hausdorff and K is compact in its subspace topology.U_y,V_y — separating neighborhoodsm — finite-subcover sizemko_definition · cited — For each x outside K and each y in K, choose disjoint open U_y containing x and V_y containing y. The V_y cover K, so compactness supplies a finite subcover. The corresponding finite intersection of U_y is an open neighborhood of x disjoint from K; therefore X minus K is open. Empty K is closed too. The Hausdorff hypothesis cannot be removed, and closedness does not conversely imply compactness in general. · mko-compact-subset-closed-in-hausdorffsets · 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.hausdorff_def · prerequisite — Hausdorffness: every distinct pair has disjoint open neighborhoods containing the assigned points.compact_def · prerequisite — K is compact when every relative-open cover has a finite original-index subcover; relative opens are intersections of ambient opens with K.Before
K=\varnothing
After
X\setminus K=X\text{ is open}Conditions: none
Rule: sets · Uses: G1
Leaf basis: sets
Handle empty K first: its complement is the open whole space.
Before
K\ne\varnothing,\quad x\in X\setminus K
After
x\ne y\quad\forall y\in K
Conditions: K\ne\varnothing
Rule: sets · Uses: G1
Leaf basis: sets
Fix an arbitrary outside point x; it differs from every y in K.
Before
x\ne y\in K\cup\{x\}After
x\in U_y,\ y\in V_y,\ U_y,V_y\text{ open},\ U_y\cap V_y=\varnothingConditions: none
Rule: hausdorff_def · Uses: G1, S02, hausdorff_def
Leaf basis: hausdorff_def
For each y use Hausdorff separation, retaining both neighborhoods.
Before
y\in K\cap V_y\quad(y\in K)
After
K\subseteq\bigcup_{y\in K}(K\cap V_y)Conditions: none
Rule: sets · Uses: S03
Leaf basis: sets
Each y lies in its own V_y, and intersection with K is relatively open.
Before
(K\cap V_y)_{y\in K}\text{ covers }KAfter
K\subseteq\bigcup_{j=1}^{m}V_{y_j},\quad m\in\mathbb NConditions: K\ne\varnothing
Rule: compact_def · Uses: G1, S04, compact_def
Leaf basis: compact_def
Compactness selects finitely many original indices; K nonempty permits m≥1.
Before
U=\bigcap_{j=1}^{m}U_{y_j}After
U\text{ is open and }x\in UConditions: m<\infty
Rule: topology · Uses: S03, S05
Leaf basis: topology
Finite intersections of opens are open, and every factor contains x.
Before
z\in U\cap K
After
z\in V_{y_j}\cap U_{y_j}=\varnothing\text{ for some }jConditions: none
Rule: sets · Uses: S03, S05, S06
Leaf basis: sets
A point z in U and K lies in some selected V, while U lies in the corresponding U-neighborhood, a contradiction.
Before
\forall x\in X\setminus K\ \exists U_x\text{ open}:x\in U_x\subseteq X\setminus KAfter
X\setminus K=\bigcup_{x\notin K}U_x\text{ open};\quad K\text{ closed}Conditions: none
Rule: topology · Uses: S01, S02, S06, S07
Leaf basis: topology
The complement is the union of these open neighborhoods, hence open; an empty complement is the empty open union.
No substeps in this guide.
S08 · premise_audit · S01, S02, S06, S07 — This establishes only the forward theorem, not that every closed set is compact.S07 · premise_audit · S03, S05, S06 — The index j may depend on z; no single V is assumed to cover K.S06 · premise_audit · S03, S05 — The finite-intersection axiom, not arbitrary intersection closure, is used.S05 · premise_audit · G1, S04, compact_def — Finiteness is the essential bridge to openness of the next intersection.S04 · premise_audit · S03 — No claim is made yet that the intersection of every U_y is open; that could be infinite.S03 · premise_audit · G1, S02, hausdorff_def — Neighborhoods for one y alone need not cover K.S02 · premise_audit · G1 — Arbitrariness of x will establish openness of the whole complement.S01 · premise_audit · G1 — This branch requires no point in K or nonempty subcover.Select finitely many V's before intersecting the corresponding U's.
Separate a fixed outside point from every point of K, then release the outside-point quantifier.
reject — A singleton in the indiscrete two-point space is compact but not closed.reject — Only finite intersections are guaranteed open; compactness must be used first.reject — R is closed; the cover (-n,n), n≥1, has no finite subcover.Under the two stated hypotheses K is closed, including empty K.
0 declared checks for this guide; only the listed expressions are covered.
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