# Continuous function from Compact Metric Space to Metric Space is Uniformly Continuous

In Rudin 4.10 he wants us to show that a continuous function from a compact metric space to a metric space is uniformly continuous by deriving that if $$f$$ is not uniformly continuous then there is an $$\epsilon >0$$ such that there are two sequences $$(p_n)$$ and $$(q_n)$$ such that $$d(p_n, q_n) \to 0$$ and $$d(f(p_n,f(q_n))>\epsilon$$ and then, using the fact that every infinite subset of a compact metric space has a limit point in that space, get a contradiction.

### Proof

$$f$$ not uniformly continuous on $$X$$ implies that $$\exists \epsilon >0 : \forall \delta >0, x \in X: \exists x'\in X : d(x,x')<\delta$$ but $$d(f(x),f(x'))>\epsilon$$

So let $$p_{n}\rightarrow x$$ and $$q_{n}\rightarrow x'$$

(Need to show existence?)

Since $$\{p_n\}$$ and $$\{q_n\}$$ are infinite subsets of $$f(X)$$, a compact set by continuity, both of them have their limits in $$f(X)$$ but that would mean $$f$$ is discontinuous at $$x$$, as two sequences are converging to different values at $$x$$.

Is this the contradiction that they're looking for? Thanks.

You are correct. To make it precise, first assume that $f$ is not uniformly continuous, then there is a fixed $\epsilon >0$ and $p_n, q_n$ such that
$$d(p_n, q_n) \leq 1/n \ \text{ and }d(f(p_n)), f(q_n) ) > \epsilon$$
By passing to subsequence if necessary, we can assume that $p_n \to x$ as $X$ is compact. As $d(p_n, q_n) \to 0$, we also have $q_n \to x$. By continuity of $f$,
$$\lim_{n\to \infty} f(p_n) = f(x) = \lim_{n\to \infty} f(p_n) .$$
But this is a impossible as $d(f(p_n), f(q_n)) > \epsilon$. Thus no such two sequences could be found and $f$ is uniformly continuous.