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

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

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Extra stuff, structure, properties

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topological homotopy theory

Analysis

Contents

Statement

Proposition

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

Let KK and YY be two metric spaces regarded as topological spaces via their metric topology, such that KK compact.

Then every continuous function of the form

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

is uniformly continuous.

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

Proof

Assume on the contrary that it were not. This would mean that for all ϵ(0,)\epsilon \in (0,\infty) there would for all nn \in \mathbb{N} be points x n,y nKx_n,y_n \in K with d K(x n,y n)<1/(n+1)d_K(x_n,y_n) \lt 1/(n+1) but d Y(f(x n),f(y n))>ϵd_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(x_{n_k})_{k \in \mathbb{N}}. Since d K(x n ky n k)<1/(n+1)d_K(x_{n_k} y_{n_k}) \lt 1/(n+1) it follows that also (y n k) k(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

limkf(x n k)=limkf(y n k). \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))>ϵd_Y(f(x_k), f(y_k)) \gt \epsilon. Hence we have a proof by contradiction.

Revised on June 21, 2017 04:48:49 by Urs Schreiber (131.220.184.222)