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I need to prove that an isometry $f$ on a compact metric space $X$ is necessarily bijective. I've got most of the proof, but I can't figure out why any point in $X-f(X)$ would necessarily have to have some open neighborhood disjoint from $f(X)$.

yrudoy
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    Spoiler found by Google (first hit on searching the title of the question): http://at.yorku.ca/cgi-bin/bbqa?forum=homework_help_2003;task=show_msg;msg=0875.0001 – lhf Apr 21 '11 at 17:39
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    In other words, you want to know why the image of the compact set $X$ under the continuous map $f$ is closed in the metric space $X$? – Jonas Meyer Apr 21 '11 at 17:42
  • Aha, no need to post it. The proof is standard. –  Apr 21 '11 at 17:44
  • @lhf Unfortunately, it just states "Take x in $X \ f(X), 0 < epsilon < dist(x,f(X))$". Why is it that $dist(x,f(X))>0$? – yrudoy Apr 21 '11 at 17:45
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    @yrudoy: Have you seen any theorems involving compact spaces and continuous maps, or involving compact subsets of metric (or Hausdorff) spaces? – Jonas Meyer Apr 21 '11 at 17:47
  • @Jonas: Sure. But I can't think of one that would guarantee it, though I wouldn't be surprised if I was overlooking something. – yrudoy Apr 21 '11 at 17:50
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    @yrudoy: A compact subset of a Hausdorff space is closed. i.e. the complement is open... –  Apr 21 '11 at 17:51
  • @yrudoy: $f$ is an isometry. You should be able to show that it is uniformly continuous using the definition. – Jonas Meyer Apr 21 '11 at 18:07
  • got it. I was using the topological definition of continuity and it isn't immediately apparent from that angle. – yrudoy Apr 21 '11 at 18:12
  • BTW, here is a sort of converse: http://math.stackexchange.com/questions/12285/isometry-in-compact-metric-spaces – lhf Apr 21 '11 at 20:23

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$f(X)$ is compact. If $x_0\notin f(X)$, because $X$ is separated for all $x\in f(X)$ exists two disjoints open sets $U_x$ and $V_x$ such that $x_0\in U_x$ and $x\in V_x$. We can find $n\in\mathbb N$ and $x_1,\cdots,x_n\in f(X)$ such that $f(X)\subset \bigcup_{j=1}^nV_{x_j}$. Now put $U:=\bigcap_{j=1}^nU_{x_j}$.

Davide Giraudo
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