TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones

Statement

Let $(X, T)$ be a Hausdorff topological space (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), with compact subsets as in Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right. Then:

A point and a disjoint compact set are separated. If $K \subseteq X$ is compact and $x \in X ∖ K$ , there are $U, V \in T$ with $x \in U, K \subseteq V, U \cap V = \emptyset .$
Two disjoint compact sets are separated. If $K, L \subseteq X$ are compact and $K \cap L = \emptyset$ , there are $U, V \in T$ with $L \subseteq U, K \subseteq V, U \cap V = \emptyset .$
Compact implies closed. Every compact subset of $X$ is closed in $X$ .
In a compact Hausdorff space the two classes coincide. If in addition $(X, T)$ is compact, then a subset of $X$ is compact if and only if it is closed.

The proof is written choice-free, and that is not a stylistic preference. The textbook argument says "for each $y \in K$ choose disjoint open $U_{y}, V_{y}$ ", which is a selection over an arbitrary index set and therefore an appeal to the full Axiom of Choice. What is done below instead is to take the family of all open $V$ that admit some open $U ∋ x$ disjoint from them — a family cut out by a formula, with nothing selected — extract a finite subcover from it, and only then make finitely many selections, which Every natural-number-indexed list of nonempty sets has a choice function on its family of values supplies as a theorem of ZF.

Facts & Assumptions

Given: A Hausdorff topological space $(X, T)$ .

[A1]

For all $x, y \in X$ with $x \neq y$ there are $U, V \in T$ with $x \in U$ , $y \in V$ and $U \cap V = \emptyset$ (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L1]

$\emptyset$ and $X$ are open, an arbitrary union of open sets is open, the intersection of finitely many open sets is open when at least one is taken, and a subset is closed exactly when its complement is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

A subset $A \subseteq X$ is a compact subset of $X$ exactly when for every family $U$ of open subsets of $X$ with $A \subseteq ⋃ U$ there are $n \in N$ and $U_{0}, \dots, U_{n} \in U$ with $A \subseteq U_{0} \cup \dots \cup U_{n}$ , or else $A = \emptyset$ (A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, claim 1; Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L3]

A function with domain a natural number all of whose values are nonempty sets has a choice function, and this is a theorem of ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[L4]

A closed subset of a compact space is a compact subset of it (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, claim 1).

Proof

technique · direct

1.1construct

For claim 1 fix a compact $K \subseteq X$ and a point $x \in X ∖ K$ , and put $V : = {V \in T : U \cap V = \emptyset for some U \in T with x \in U}$ , a family cut out by a property of $V$ alone and not by any selection.

2.1A1step 1.1

$K \subseteq ⋃ V$ : given $y \in K$ we have $y \neq x$ , since $x \notin K$ , so [A1] provides $U, V \in T$ with $x \in U$ , $y \in V$ and $U \cap V = \emptyset$ ; that $V$ belongs to $V$ and contains $y$ .

3.1L1L2step 1.1step 2.1

If $K = \emptyset$ then $U : = X$ and $V : = \emptyset$ satisfy claim 1; otherwise [L2] applied to the family $V$ gives $n \in N$ and $V_{0}, \dots, V_{n} \in V$ with $K \subseteq V_{0} \cup \dots \cup V_{n}$ .

4.1L3step 3.1

For each $j \leq n$ the set $S_{j} : = {U \in T : x \in U and U \cap V_{j} = \emptyset}$ is nonempty, because $V_{j} \in V$ ; and $j \mapsto S_{j}$ is a function with domain the natural number $σ (n)$ , so a choice function for its values supplies $U_{0}, \dots, U_{n} \in T$ with $x \in U_{j}$ and $U_{j} \cap V_{j} = \emptyset$ for every $j \leq n$ .

5.1L1step 3.1step 4.1

Put $U : = U_{0} \cap \dots \cap U_{n}$ and $V : = V_{0} \cup \dots \cup V_{n}$ ; both are open by [L1], $x \in U$ because $x \in U_{j}$ for every $j$ , $K \subseteq V$ by step 3.1, and $U \cap V = \emptyset$ because a point of $U \cap V$ would lie in some $V_{j}$ and in $U \subseteq U_{j}$ , contradicting $U_{j} \cap V_{j} = \emptyset$ . So claim 1 holds.

6.1L1step 5.1

For claim 3 let $K \subseteq X$ be compact and put $G : = ⋃ {W \in T : W \cap K = \emptyset}$ , which is open by [L1]. Every member of the union misses $K$ , so $G \subseteq X ∖ K$ ; conversely for $x \in X ∖ K$ claim 1, proved at step 5.1, gives disjoint open $U ∋ x$ and $V \supseteq K$ , whence $U \cap K = \emptyset$ and $x \in U \subseteq G$ . So $G = X ∖ K$ is open, $K$ is closed, and claim 3 holds.

6.2step 5.1construct

For claim 2 let $K, L \subseteq X$ be compact with $K \cap L = \emptyset$ , and put $W : = {W \in T : V \cap W = \emptyset for some V \in T with K \subseteq V}$ , again cut out by a property. Then $L \subseteq ⋃ W$ : for $y \in L$ we have $y \notin K$ , so claim 1, proved at step 5.1, gives disjoint open $U ∋ y$ and $V \supseteq K$ , and that $U$ lies in $W$ and contains $y$ .

7.1L1L2step 6.2

If $L = \emptyset$ then $U : = \emptyset$ and $V : = X$ satisfy claim 2; otherwise [L2] applied to $W$ gives $m \in N$ and $W_{0}, \dots, W_{m} \in W$ with $L \subseteq W_{0} \cup \dots \cup W_{m}$ .

8.1L3step 7.1

For each $j \leq m$ the set $T_{j} : = {V \in T : K \subseteq V and V \cap W_{j} = \emptyset}$ is nonempty, because $W_{j} \in W$ ; and $j \mapsto T_{j}$ is a function with domain the natural number $σ (m)$ , so a choice function for its values supplies $V_{0}, \dots, V_{m} \in T$ with $K \subseteq V_{j}$ and $V_{j} \cap W_{j} = \emptyset$ for every $j \leq m$ .

9.1L1step 7.1step 8.1

Put $U : = W_{0} \cup \dots \cup W_{m}$ and $V : = V_{0} \cap \dots \cap V_{m}$ ; both are open by [L1], $L \subseteq U$ by step 7.1, $K \subseteq V$ because $K \subseteq V_{j}$ for every $j$ , and $U \cap V = \emptyset$ because a point of $U \cap V$ would lie in some $W_{j}$ and in $V \subseteq V_{j}$ , contradicting $V_{j} \cap W_{j} = \emptyset$ . So claim 2 holds.

10.1L4step 6.1step 9.1∎

For claim 4 assume $(X, T)$ is also compact: a compact subset of $X$ is closed by step 6.1, and a closed subset of $X$ is compact by [L4], so the two classes of subsets coincide; with claims 1, 2 and 3 settled at steps 5.1, 9.1 and 6.1 the theorem is proved.

Remarks

Where each hypothesis is spent. The Hausdorff condition is used exactly once, at step 2.1, to know that the family $V$ covers $K$ ; compactness of $K$ is used exactly once, at step 3.1, to cut that cover down to finitely many members. Claim 2 then reuses claim 1 in the same shape, with the roles of point and compact set played by a point of $L$ and the compact set $K$ .

Why the family is defined and not chosen. For each $y \in K$ the Hausdorff condition asserts that some pair $(U, V)$ exists; it provides no rule for naming one. A proof that writes $U_{y}$ and $V_{y}$ has selected a pair for every $y \in K$ at once, and for an arbitrary compact $K$ that is the Axiom of Choice. Collecting instead every $V$ that works for some $U$ replaces the selection by a formula, and the only selection left is over the finite index set $σ (n)$ , where Every natural-number-indexed list of nonempty sets has a choice function on its family of values applies.

Claim 3 fails without the Hausdorff hypothesis, and FALSE: a compact subset of a topological space is closed records the failure with a witness. Claim 4 is the converse pairing: closedness is enough for compactness only when the ambient space is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact), and compactness is enough for closedness only when it is Hausdorff.

In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources