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[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!

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    $\begingroup$ In inequality with functions you should have more or less $\geqslant \epsilon_0$. $\endgroup$ – zkutch Sep 1 '20 at 2:16
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    $\begingroup$ @EvanWilliamChandra That can go as an answer below, it is that good an explanation. $\endgroup$ – Teresa Lisbon Sep 1 '20 at 2:48
  • $\begingroup$ @Theresa Lisbon Thank you for your comment. I'll move my comment above to the answer section then. $\endgroup$ – Evan William Chandra Sep 1 '20 at 3:18
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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.

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  • $\begingroup$ +1... To a student I would say: Take an infinite $S\subset \Bbb N$ such that $(x_n)_{n\in S}\to x.$ And take an infinite $T\subset S$ such that $(y_n)_{n\in T}\;$...(which is a subsequence of $(y_n)_{n\in S})$... converges to $y.$ And I would emphasize that $x,y \in K$ as $K$ must be closed. $\endgroup$ – DanielWainfleet Sep 1 '20 at 6:05

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