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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)
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The meagre subsets of a topological space form a sigma-ideal

Statement

For every topological space X, the meagre subsets of X contain ∅ and are closed under taking subsets; assuming the Axiom of Countable Choice, they are also closed under countable unions. Countable Choice is what selects one witnessing sequence of nowhere dense sets for each member of the countable family, before the flattening bijection is applied.

Facts & Assumptions

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

[F1]

Let X be a topological space and let A⊆X. The set A is nowhere dense when int⁡(A‾)=∅ (def-interior-closure-boundary-top). It is meagre when there is a sequence (Nn)n∈N of nowhere dense subsets of X with A⊆⋃nNn. It is residual, or comeagre, when X∖A is meagre. The empty union shows that ∅ is meagre, including when X=∅. (Nowhere dense, meagre, residual, and comeagre subsets of a topological space).

[F2]

N×N≈N (def-equinumerous): the plane of pairs of naturals is countably infinite (def-countable). The bijection is exhibited, not merely asserted to exist. Define 2m by recursion on m (thm-recursion) by 20=1 and 2σ(m)=2m+2m, and set J(m,n)=2m⋅σ(n+n),that isJ(m,n)=2m(2n+1). Then J is a bijection from N×N onto N∖{0}, and σ is a bijection from N onto N∖{0}, so σ−1∘J is a bijection N×N→N. What makes J bijective is the decomposition of a nonzero natural into a power of two times an odd number, existence and uniqueness both. (N×N≈N).

[A1]

The Axiom of Countable Choice (ACω) selects one member from each nonempty set in a sequence of sets. It is used below for the sets of nowhere dense covering sequences.

Proof

technique · direct
1.1givenF1

The empty set is meagre, witnessed by the constant sequence of empty nowhere dense sets. If B⊆A and (Nn) witnesses that A is meagre, the same sequence witnesses that B is meagre.

1.2givenF1A1

Let (Am)m∈N be meagre. For each m, let Wm be the nonempty set of sequences (Nm,n)n∈N of nowhere dense subsets of X satisfying Am⊆⋃nNm,n. Apply [A1] once to (Wm) to choose all these sequences simultaneously. This is the only use of Countable Choice.

2.1F1F2step 1.2

Let b:N→N×N be the bijection in [F2], and put Mk=Nb(k). Every Mk is nowhere dense, and ⋃mAm⊆⋃kMk. Thus the countable union is meagre; the empty indexed family has union ∅ as in step 1.1.

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the stated sigma-ideal properties under the stated choice hypothesis.

Depends on

Used by

Dependency tree · two levels

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Sources