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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.


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

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 $d(f(x_{n_k},f(y_{n_k})\geq c$

  • $\begingroup$ What is the definition of a compact mapping? $\endgroup$ – user99914 Jun 11 '16 at 23:21
  • 1
    $\begingroup$ 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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