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Let $(M,d)$ be a compact metric space and $f: M \rightarrow M$ a continuous function. I'm trying to prove that if $d(f(x),f(y)) \geq d(x,y)$ for every $x,y \in M$, then $f$ is an isometry. This is how far I could get: Since $f$ is a continuous function, the set $\{(x,y) \in M \times M : d(f(x),f(y)) = d(x,y)\}$ is closed, and hence compact. By symmetry, it is of the form $N \times N$. Thus, $N$ (with the metric induced by $M$) is a compact metric space such that $d(f(x),f(y)) = d(x,y)$ for every $x,y \in N$. I already know that this implies that the restriction of $f$ to $N$ is an isometry of $N$, but how to prove that the open set $M \setminus N$ is empty?

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Hint: suppose $x,y\in M$ are such that $d(fx,fy)>d(x,y)$. By induction, get a sequence of points $z_n$ such that $d(z_n,y)>d(z_{n-1},y)$ for all $n$. Combine with compactness of $M$. –  wildildildlife Jul 14 '11 at 13:06
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A good exercise as a follow-up: prove that $f$ must also be surjective. –  Mark Jul 14 '11 at 16:18

1 Answer 1

up vote 2 down vote accepted

Look at the 12th post. Here

(This post is actually not correct and/or incomplete. The subsequences of a_n and b_n which converge due to compactness of the space do not necessarily have the same sub-indices, and so we cannot necessarily use the same function g(n) for both subsequences. The idea is in the right direction, but this does not suffice as it is)

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that's a nice argument (the one you linked) –  Zarrax Jul 14 '11 at 14:50
    
You can find a common subsequence: Find $h(n)$ so that $a_{h(n)}$ converges. The sequence $b_{h(n)}$ has a convergent subsequence, which can be defined by the function $g(n)$, and $a_{g(n)}$ converges since it is a subsequence of a convergent sequence. –  yasmar Feb 21 '12 at 7:58
    
It seems everything is beyond my reach! Let me ask the Community why the added paragraph was not written as a comment first. That was a very naive way of down voting, also. –  Ehsan M. Kermani Feb 21 '12 at 9:32

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