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Theorem If $f$ is continuous on a compact metric space $X$, then $f$ is uniformly continuous on $X$.

Proof Let $\epsilon>0$. For any $y\in X$ there is a $\delta_y$ such that $d(x,y)<\delta_y$ implies $d(f(x),f(y))<\epsilon$. Now, let $\bigcup{\mathcal{U_y}}$ be an open cover of $X$ where $\mathcal{U}_y=B(y;\delta_y)$. Since $X$ is compact $\bigcup{\mathcal{U_y}}$ has a Lebesgue number $\delta$. Let $A\subset X$ such that for all $x,y\in A$, $d(x,y)<\delta$. Thus, as $A\subset B(y;\delta_y)$ for some $y$, we have $d(x,y)<\delta_y$, which implies $d(f(x),f(y))<\epsilon$.

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    It looks ok, take a look at this (Its a bit neater) : http://mathramble.wordpress.com/2013/03/12/the-lebesgue-number-lemma-and-uniform-continuity/ – user10444 Apr 21 '13 at 13:50

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