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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)
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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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Open subspaces and residual subspaces of Baire spaces are Baire

Statement

Every open subspace of a Baire space is Baire. Every residual subspace of a Baire space, with its subspace topology, is Baire.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

For a topological space X, the following are equivalent: every countable intersection of dense open sets is dense; every countable union of closed sets with empty interior has empty interior; no nonempty open subset is meagre in X; and every residual subset meets every nonempty open set. The equivalence includes the empty space. (Equivalent forms of the Baire property).

[F2]

A set is residual when its complement is contained in the union of one sequence of nowhere dense sets (Nowhere dense, meagre, residual, and comeagre subsets of a topological space).

[F3]

Let (X,T) be a topological space (def-topological-space) and let S⊆X. The subspace topology (also relative topology) on S is TS:={ U∩S:U∈T }, the family of traces on S of the open sets of X. The pair (S,TS) is a subspace of X. A subset of S that lies in TS is said to be open in S, and relatively open where the ambient space needs emphasis. (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).

Proof

technique · direct
1.1F1F3given

Let O be open in a Baire space X, and let (Gn) be dense open subsets of O. Put Fn=O∖Gn and Cn=Fn‾X. Each Cn is nowhere dense in X: if a nonempty X-open V lay in Cn, then V would meet O (since Cn⊆O‾), while the nonempty relatively open V∩O would lie in Cn∩O=Fn, contradicting density of Gn in O.

1.2F1F2given

Let Y be residual in X. By [F2], fix one sequence (Nn) of nowhere dense subsets with X∖Y⊆⋃nNn. Since no nonempty open subset of a Baire space is meagre [F1], Y is dense in X.

2.1F1step 1.1

The sets X∖Cn are dense open in X. By the Baire property, their intersection meets every nonempty open subset V of O; a point in that intersection and V lies in every Gn. Thus O is Baire, including the empty case.

2.2F3step 1.2

Let (Gn) be dense open subsets of Y. Put Fn=Y∖Gn and Cn=Fn‾X. Because Y is dense, each Cn is nowhere dense in X: otherwise a nonempty X-open V⊆Cn would meet Y, and the nonempty relatively open V∩Y would lie in Cn∩Y=Fn, contradicting density of Gn in Y.

3.1F1F2F3step 1.2step 2.2

The single interleaved sequence N0,C0,N1,C1,… witnesses that (X∖Y)∪⋃nCn is meagre; forming it requires no countable selection of witnesses. Given a nonempty relatively open W=V∩Y in Y, V is nonempty open in X. By [F1], V is not contained in that meagre set. Any point of V outside it belongs to Y and every Gn, hence to W∩⋂nGn. Therefore Y is Baire.

4.1step 2.1step 3.1∎

Steps 2.1 and 3.1 establish both assertions.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources