Contents

# Contents

## Statement

###### Proposition

(continuous metric space valued function on compact metric space is uniformly continuous)

Let $K$ and $Y$ be two metric spaces regarded as topological spaces via their metric topology, such that $K$ compact.

Then every continuous function of the form

$K \overset{f}{\longrightarrow} Y$

In particular since subspaces of metric spaces canonically are metric spaces, it follows that if $X$ is a metric space

###### Proof

Assume on the contrary that it were not. This would mean that for all $\epsilon \in (0,\infty)$ there would for all $n \in \mathbb{N}$ be points $x_n,y_n \in K$ with $d_K(x_n,y_n) \lt 1/(n+1)$ but $d_Y(f(x_n),f(y_n)) \gt \epsilon$.

Since in compact metric spaces every sequence has a converging subsequence (here), it follows that there is a converging subsequences $(x_{n_k})_{k \in \mathbb{N}}$. Since $d_K(x_{n_k} y_{n_k}) \lt 1/(n+1)$ it follows that also $(y_{n_k})_{k \in \mathbb{N}}$ is a converging subsequence which converges to the same point:

But since continuous functions preserve limits of sequences, it would follow that

$\underset{k \to \infty}{\lim} f(x_{n_k}) \;=\; \underset{k \to \infty}{\lim} f(y_{n_k}) \,.$

This however would contradict the assumption that $d_Y(f(x_k), f(y_k)) \gt \epsilon$. Hence we have a proof by contradiction.

Last revised on June 21, 2017 at 08:48:49. See the history of this page for a list of all contributions to it.