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 $f(X)$.
isometry on compact space
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general-topology
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2Spoiler 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 – 2011-04-21
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2In 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 $X$? – 2011-04-21
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0Aha, no need to post it. The proof is standard. – 2011-04-21
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0@lhf Unfortunately, it just states "Take x in $X \ f(X), 0 < epsilon < dist(x,f(X))$". Why is it that $dist(x,f(X))>0$? – 2011-04-21
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1@yrudoy: Have you seen any theorems involving compact spaces and continuous maps, or involving compact subsets of metric (or Hausdorff) spaces? – 2011-04-21
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0@Jonas: Sure. But I can't think of one that would guarantee it, though I wouldn't be surprised if I was overlooking something. – 2011-04-21
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1@yrudoy: A compact subset of a Hausdorff space is closed. i.e. the complement is open... – 2011-04-21
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0@yrudoy: $f$ is an isometry. You should be able to show that it is uniformly continuous using the definition. – 2011-04-21
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0got it. I was using the topological definition of continuity and it isn't immediately apparent from that angle. – 2011-04-21
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0BTW, here is a sort of converse: http://math.stackexchange.com/questions/12285/isometry-in-compact-metric-spaces – 2011-04-21
1 Answers
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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 $U:=\bigcap_{j=1}^nU_{x_j}$.
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0What is the contradiction? Where do you use the isometry? – 2011-04-21
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1@Henno: this is an answer to the question "I can't figure out why..." (i.e. just a proof that $f(X)$ is closed). – 2011-04-21
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0I red too fast the problem. I tought that Henno wanted a neighborhood of $x_0$ disjoint form a neighborhood of $f(X)$. – 2011-04-21