ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

A finite subset of any space is compact, so the compact separation clauses specialise to separating a point from a finite set in a Hausdorff space

Example

Let $X$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $F \subseteq X$ be finite (Finite, countably infinite, countable, uncountable), with the subspace topology (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). Then:

$F$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), whatever $X$ is and whatever topology it carries.
Consequently, if $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) then a point $x \in X \setminus F$ and the set $F$ have disjoint open neighbourhoods, and two disjoint finite subsets of $X$ have disjoint open neighbourhoods (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); in particular $F$ is closed in $X$ .

Clause 1 spends a choice principle, and exactly one: finite choice (Every natural-number-indexed list of nonempty sets has a choice function on its family of values), which is a theorem of ZF. The naive phrasing of the same argument — "for each $y \in F$ pick a member of the cover containing it" — is a selection over the index set of $F$ , and because that index set is a natural number the selection is licensed outright.

Facts & Assumptions

Given: A topological space $X$ , a finite subset $F \subseteq X$ with the subspace topology, and, where clause 2 is at issue, the hypothesis that $X$ is Hausdorff.

[A1]

$F$ is finite, so $F$ is equinumerous with a natural number $n$ and may be listed as $y_0, \dots, y_{n-1}$ (Finite, countably infinite, countable, uncountable).

[A2]

A space is compact when every family of its open sets whose union is the whole space has a finite subfamily whose union is the whole space; a subset is compact when it is compact as a subspace (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).

[L1]

If $G$ is a function with domain a natural number $n$ all of whose values are nonempty sets, then the family of its values has a choice function; this is a theorem of ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values, Choice function).

[L2]

In a Hausdorff space a point and a disjoint compact set have disjoint open neighbourhoods, two disjoint compact sets have disjoint open neighbourhoods, and every compact subset is closed (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, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

Verification

technique · direct

1.1

List $F$ as $y_0, \dots, y_{n-1}$ for a natural number $n$ , and let $\mathcal{U}$ be a family of sets open in the subspace $F$ whose union is $F$ .

A1A2

2.1

For each $i < n$ the set $\mathcal{U}_i := \{\, O \in \mathcal{U} : y_i \in O \,\}$ is nonempty, since the union of $\mathcal{U}$ is $F$ and $y_i \in F$ ; so by [L1] applied to the function $i \mapsto \mathcal{U}_i$ on $n$ there is a choice function on the family of these sets, and it supplies $O_i \in \mathcal{U}_i$ for every $i < n$ .

step 1.1L1choose

3.1

The finitely many sets $O_0, \dots, O_{n-1}$ lie in $\mathcal{U}$ and their union contains every $y_i$ , hence is $F$ ; as $\mathcal{U}$ was arbitrary, $F$ is compact, which is claim 1.

step 1.1step 2.1A2

4.1

If $X$ is Hausdorff then, $F$ being compact by step 3.1, [L2] separates $F$ from any point of $X \setminus F$ by disjoint open sets, separates $F$ from any disjoint finite subset of $X$ likewise, and makes $F$ closed in $X$ . This is claim 2.

step 3.1L2∎

Remarks

Clause 2 recovers the $T_1$ behaviour of a Hausdorff space by a different route. That every finite subset of a Hausdorff space is closed is usually read off from the separation axioms; here it arrives as a special case of a compactness statement, and the two readings agree, as they must.
Where the finiteness of $F$ is used. Only in step 2.1, and only to make the selection a finite one. The same argument with an infinite $F$ would need a genuine choice principle and would in any case fail at step 3.1, an infinite index set producing no finite subcover.
The example is the smallest non-trivial instance of the compact separation clauses. It needs no compactness hypothesis on $X$ and no cover argument beyond the one above, so it is the case in which the clauses of 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 can be checked against intuition before being used on genuinely compact sets.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Compact space (Wikipedia) (standard reference, not scraped)
Hausdorff space (Wikipedia) (standard reference, not scraped)
A. Hatcher, Topology Notes (standard reference, not scraped)