Bourbaki

Corollary 1

Tag 0BD8, permanent. Cite it as [Bourbaki, Tag 0BD8].

General Topology, chapter II, § 1. Uniform spaces

If AA is any subset of XX and VV is any symmetric entourage of XX, then V(A)V(A) is a neighbourhood of AA in XX, and

A‾=⋂V∈SV(A)=⋂U∈UV(A) \overline{A} = \bigcap_{V \in \mathscr{S}} V(A) = \bigcap_{U \in \mathfrak{U}} V(A)

where U\mathfrak{U} denotes the set of all entourages in XX.

If M=A×AM = A \times A, then VMV=V(A)×V(A)VMV = V(A) \times V(A) for any V∈SV \in \mathscr{S}; for the relation "there exists p∈Ap \in A such that (x,p)∈V(x, p) \in V" is by definition equivalent to x∈V(A)x \in V(A). The corollary now follows from Chapter I, § 4, no. 2, Proposition 5 and no. 3, Proposition 7.

V(A)V(A) is said to be the V-neighbourhood of AA.

If VV is an entourage which is open in X×XX \times X, then V(x)V(x) is open in XX for each x∈Xx \in X (Chapter I, § 4, no. 2, Corollary to Proposition 4) and therefore V(A)V(A), being the union of the V(x)V(x) as xx runs through AA, is open in XX. On the other hand, if VV is a closed entourage in X×XX \times X, V(A)V(A) need not be closed in XX for every subset AA of XX (Exercise 3).

It should also be remarked that as VV runs through the set of entourages of XX, the sets V(A)V(A) do not necessarily form a fundamental system of neighbourhoods of AA in XX (Exercise 2).

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

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