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This question already has an answer here:

Why is a continuous mapping from a compact metric space to another metric space is uniformly continuous? This theorem is from Rudin Real Analysis Page 202.

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marked as duplicate by user99914, user223391, Daniel W. Farlow, user228113, M. Vinay Jun 12 '16 at 4:11

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  • $\begingroup$ What is the definition of compact mapping? $\endgroup$ – user99914 Jun 11 '16 at 23:16
  • $\begingroup$ In particular, I cannot find your theorem in Rudin "real analysis" (all three of them). Can you state precisely the reference, including the edition? $\endgroup$ – user99914 Jun 11 '16 at 23:26
  • $\begingroup$ Sorry, typo. It should be continuous mapping. Updated the question. $\endgroup$ – user1559897 Jun 11 '16 at 23:37
  • $\begingroup$ It goes by the name of Heine-Cantor theorem. $\endgroup$ – user228113 Jun 12 '16 at 3:51
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Let $f:X\rightarrow Y$ be a continuous map defined on metric spaces such that $X$ is compact. Suppose it is not uniformly continuous, there exists $c>0$ such that for every integer $n$, there exists $x_n,y_n$ such that $d(x_n,y_n)<1/n$ and $d(f(x_n),f(y_n))>c$, you can extract a subsequence $(x_{n_k})$ from $(x_n)$ which converges towards $x$, $(y_{n_k})$ converges towards $x$, this implies that $lim_nd(f(x_{n_k}),f(y_{n_k})=0$. This is impossible since $d(f(x_{n_k},f(y_{n_k})\geq c$

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  • $\begingroup$ What is the definition of a compact mapping? $\endgroup$ – user99914 Jun 11 '16 at 23:21
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    $\begingroup$ I believe a continuous map $f:X\rightarrow Y$ such that $X$ is compact $\endgroup$ – Tsemo Aristide Jun 11 '16 at 23:23

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