Prove by reduction to the absurd that if $K$ is compact and $f$ is continuous on $K$ then $f$ is uniformly continuous in $K$.

Question

[Heine–Cantor theorem] Prove by reduction to the absurd that if $K\subset\mathbb{R^m}$ is compact and $f: K\longrightarrow \mathbb{R}$ is continuous on $K$ then $f$ is uniformly continuous in $K$.

I know the proof by useing balls but I'm haveing a little trouble doing it by reduction to the absurd.

I'm trying to use the definition with sequences:

$f$ is not uniformly continuous in $K$ if there exists $\epsilon_0> 0$ such that for all $n \in\mathbb{N}$ there exist $x_n, y_n\in K$ such that $|x_n- y_n|<\frac{1}{n}\quad$ but $\quad|f (x_n)- f (y_n)| \geq \epsilon_0$.

Any suggestions about the process and what $x_n$ and $y_n$ could be would be great!

Answer 1

You know that $a_{n}:= x_{n}-y_{n}\to0$ as $n\to0$. Then, use compactness of $K$ to extract subsequence which converges in $K$ so that $x_{n_{k}}\to x$ and $y_{n_{k}}\to y$ as $k\to\infty$. Next, show that $x=y$ and use continuity of $f$ to see that $$\lim\limits_{k\to\infty} f(x_{n_{k}}) = f(x) = f(y) = \lim\limits_{k\to\infty} f(y_{n_{k}})$$

Finally, the rest of the argument is obvious from your contradiction assumption of $f$ being not uniformly continuous.