Kechris, p. 24

(B)

Let be a compact metrizable space and a metrizable space. We denote by the space of continuous functions from into with the topology induced by the uniform metric: where is a compatible metric for . A simple compactness argument shows that this topology is independent of the choice of .

*Proof.*

Let be two compatible metrics for . I will show that the two corresponding uniform metrics, give the same topology.

Suppose that did not give the same topology. That is, without loss of generality, there exists , , such that for all , . (Here .) That is ; that is, , which is the same as

and there exists such that.

Now, let . Let . Then there exists such that and . Since is compact, the sequence must have a convergent subsequence . So .

Let . Choose such that implies and . Then implies . So .

Therefore, converges to under . Thus, it must also do so under . But for all , . So this is a contradiction.

Hence, both metrics give the same topology.

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