Proposition 3

Tag 0C2I, permanent. Cite it as [Bourbaki, Tag 0C2I].

General Topology, chapter IX, § 2. Metric spaces and metrizable spaces

The function $d(x, A)$ is uniformly continuous on $X$ .

Let $x$ and $y$ be any two points of $X$ ; then given any $\varepsilon > 0$ there exists $z \in A$ such that $d(y, z) \leq d(y, A) + \varepsilon$ , and therefore

d(x, z) \leq d(x, y) + d(y, z) \leq d(x, y) + d(y, A) + \varepsilon

by the triangle inequality.

A fortiori $d(x, A) \leq d(x, y) + d(y, A) + \varepsilon$ , and since $\varepsilon$ is arbitrary it follows that $d(x, A) \leq d(x, y) + d(y, A)$ . Similarly we have

d(y, A) \leq d(x, y) + d(x, A),

so that

|d(x, A) - d(y, A)| \leq d(x, y),

whence the result follows.

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

Cited by

Remark, § 2
Proposition 18, § 2
Proposition 2, § 6

Cites

Nothing.