Bourbaki

Lemma 4

Tag 0C94, permanent. Cite it as [Bourbaki, Tag 0C94].

General Topology, chapter IX, § 6. Polish spaces, Souslin spaces, Borel sets

Let (An),(Am′)(A_n), (A'_m) be two sequences of subsets of a topological space X. Suppose that for each pair (An,Am′)(A_n, A'_m) there is a Borel set BnmB_{nm} of X such that Bnm⊃AnB_{nm} \supset A_n and Bnm∩Am′=∅B_{nm} \cap A'_m = \emptyset. Then there is a Borel set B of X which contains ⋃nAn\bigcup_n A_n and does not meet ⋃mAm′\bigcup_m A'_m.

For the Borel set B=⋃n(⋂mBnm)B = \bigcup_n \left( \bigcap_m B_{nm} \right) satisfies these conditions.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed volume. Not translated.