PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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.

Every compact space is paracompact

Statement

Every compact topological space is paracompact.

Facts & Assumptions

Given: A compact topological space $X$ and an open cover $U$ of $X$ .

[F1]

Compactness means that $U$ has a finite subcover (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[F2]

Paracompactness asks for a locally finite open refinement of each open cover (Paracompactness: every open cover has a locally finite open refinement, with no separation axiom built into the word).

Proof

technique · direct

1.1F1choose

By compactness, fix a finite subfamily $V \subseteq U$ covering $X$ .

2.1step 1.1

The family $V$ is open, covers $X$ , refines $U$ , and is locally finite because every point has the neighbourhood $X$ , which meets only members of the finite family $V$ .

3.1F2step 2.1∎

Thus $V$ is the refinement required by [F2], and $X$ is paracompact.

Every compact space is paracompact

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources