ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

For finite discrete $X$ and compact metric $Y$ , the whole space $C (X, Y)$ is compact

Example

Let $X$ be a finite set with the discrete topology and let $Y$ be a compact metric space. Then every map $X \to Y$ is continuous and $C (X, Y)$ is compact in the compact-open topology. This includes $X = \emptyset$ , when $C (X, Y)$ is a singleton.

Facts & Assumptions

Given: A finite discrete space $X$ and a compact metric space $Y$ .

[L1]

In the discrete topology every subset of $X$ is open (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L2]

Equicontinuity permits a neighbourhood depending on the point and tolerance but requires it to serve the whole family (Equicontinuity on a topological domain and pointwise relative compactness).

[L3]

On an equicontinuous family, the compact-open and pointwise topologies agree (The compact-open and pointwise topologies agree on an equicontinuous family).

[L4]

The pointwise topology on $Y^{X}$ is the product topology (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C (X, Y)$ ).

[L5]

Every finite product of compact spaces, including the empty product, is compact (A product of finitely many compact spaces is compact in the product topology).

Verification

technique · direct

1.1L1

Every map $f : X \to Y$ is continuous because the inverse image of each open subset of $Y$ is a subset of $X$ , hence open by [L1]. Thus $C (X, Y) = Y^{X}$ .

1.2L1L2

The whole family $Y^{X}$ is equicontinuous: at $x \in X$ , the neighbourhood ${x}$ makes $d (f (y), f (x)) = 0$ for every $f$ and every $y$ in it.

1.3L4L5

By [L4] and [L5], the pointwise topology on $Y^{X}$ is compact, including the empty product when $X = \emptyset$ .

2.1L3step 1.2step 1.3∎

By [L3], this pointwise topology equals the compact-open topology on the equicontinuous whole family. Hence $C (X, Y)$ is compact.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Topology, second edition, Section 47 (standard reference, not scraped)

For finite discrete XX and compact metric YY, the whole space C(X,Y)C(X,Y) is compact