PropositionStatement: AI-adaptedProof: AI-generatedprecheck 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.

The general compact-open topology agrees with the published metric-domain definition

Statement

Let $(X, d_{X})$ be a metric space equipped with its metric topology and let $Y$ be a topological space. The compact-open topology on $C (X, Y)$ defined using topologically compact subsets of $X$ is exactly the published compact-open topology defined using metric-compact subsets of $X$ .

Facts & Assumptions

Given: A metric space $X$ with its metric topology and a topological space $Y$ .

[L1]

The general compact-open topology has subbasis $S (K, V) = {f : f [K] \subseteq V}$ for topologically compact $K \subseteq X$ and open $V \subseteq Y$ (The compact-open topology on $C (X, Y)$ for arbitrary topological spaces).

[L2]

The published metric-domain compact-open topology has the same form of subbasis, with $K$ metric-compact (The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ ).

[L3]

A subset of a metric space is metric-compact if and only if it is compact in the metric topology (For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide).

Proof

technique · direct

1.1L3

By [L3], the compact subsets allowed in [L1] and [L2] are exactly the same subsets of $X$ .

2.1L1L2step 1.1

For every such $K$ and every open $V \subseteq Y$ , both definitions use the identical subset $S (K, V)$ of $C (X, Y)$ , including $K = \emptyset$ and $V = Y$ .

3.1step 2.1∎

The two subbasic families are equal, and therefore generate equal topologies.

Depends on

Used by

Dependency tree · two levels

19 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 46 (standard reference, not scraped)