Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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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.

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A compact subset of an open Euclidean set has a compact Jordan neighborhood inside that open set

Statement

Let n≥1. If C⊆U⊆Rn, where C is compact and U is open, then there is a compact Jordan set K such that C⊆int⁡K⊆K⊆U. The set K can be chosen as a finite union of closed grid rectangles.

Facts & Assumptions

Proof

technique · finite-cover
1.1

For each x∈C, openness gives a closed grid rectangle Rx⊆U with x∈int⁡Rx. The interiors cover C, so [L1] selects R1,…,RN. Put K=⋃iRi. Then C⊆int⁡K⊆K⊆U.

L1given
2.1

The finite union K is closed and bounded, hence compact by [L2]. Its boundary is contained in the union of the boundaries of the Ri. Each rectangular face is a continuous coordinate graph over a bounded rectangle and is null by [L3]; a finite union remains null.

L2L3step 1.1
3.1

The boundary criterion in [L3] now makes K Jordan measurable. Subdividing the finitely many rectangles by their common coordinate endpoints expresses the same set as a finite union of closed cells from one grid.

L3step 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources