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Prove that if $f$ is continuous on a compact set then it is uniformly continuous.

Proof:

Let $f:A\rightarrow \mathbb{R}$ be a continuous function and let $A$ be a compact subset in a metric space $(X,d)$.

Choose $\epsilon>0$. From continuity of $f$ we know that for every $p_i \in A$ there is $\delta_i$ such that $$|f(p_i)-f(y)|<\frac{\epsilon}{2}$$ whenever $y\in A$ and $d(p_i,y)<\delta_i$. Consider the family of open balls $$\{B(p_i,\frac{\delta_i}{2})\}\mbox{.}$$ This family is an open cover of $A$. From compactness of $A$ we can choose a finite subcover $$B(p_1,\frac{\delta_1}{2}),\ldots,B(p_n,\frac{\delta_n}{2})$$ of $A$. Now let $$\delta=\min\{\frac{\delta_1}{2},\ldots, \frac{\delta_n}{2}\} \mbox{.}$$

For any $x,y\in A$ there are $B(p_j,\frac{\delta_j}{2})$ and $B(p_k,\frac{\delta_k}{2})$ where $j,k\in\{1,\ldots, n\}$ such that $x\in B(p_j,\frac{\delta_j}{2})$ and $y\in B(p_k,\frac{\delta_k}{2})$.

If $d(x,y)<\delta$ we have $d(p_j,y)\le d(p_j,x)+d(x,y)<\delta_j$. Thus

$$|f(x)-f(y)|\le |f(x)-f(p_j)|+|f(y)-f(p_j)|<\epsilon \mbox{.}$$

My calculus is rusty and I would be grateful if someone verified my proof (I did not look it up in any book).

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    $\begingroup$ You have bad notation. The set of all $p_i$ may be uncountable, So, the finite subcover might not be as described. Instead, at the start, say "for every $p\in A$". The initial open cover is then indexed by $p\in A$ (and radii $\delta_p$). Then form the finte subcover with centers labelled $p_1,\ldots,p_n$. $\endgroup$ – David Mitra Dec 28 '15 at 9:26

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