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If $(M,d)$ is a compact metric space and $f:M→M$ is an isometry ($d(x,y)=d(f(x),f(y))$ for any $x,y∈M$), then $f$ is a homeomorphism.

I proved that $f$ is continuous as its inverse, I proved that it is injective also. I couldn't prove that it is also surjective...

thanks for any advance.

Martin Sleziak
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Armand
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    You're going to have to explain what you're asking better. Start by 1. Putting the whole question in the body, instead of spreading it across the title and body. 2. Define what exactly you mean by an isometry. 3. Use clearer language than "so that", whose meaning I do not understand in this context. – dfeuer Aug 27 '13 at 08:35
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    @dfeuer: Isometry is a standard term that needs no definition: an isometry is a distance-preserving map between metric spaces. Armand has proved that an isometry $f$ from the compact metric space $M$ to itself is a homeomorphism from $M$ onto $f[M]$ and is asking how to prove that it is surjective. – Brian M. Scott Aug 27 '13 at 08:59
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    @Armand: You’ll find a proof here. It’s fairly concise, and you may need to think about it for a bit, but it’s complete. – Brian M. Scott Aug 27 '13 at 08:59
  • @BrianM.Scott: I missed the fact that it was an autoisometry, and therefore wondered where on Earth surjectivity would come from if not in their definition. – dfeuer Aug 27 '13 at 13:58

2 Answers2

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Let $x \in X$ and $x_n=f^n(x)$. Because $X$ is compact, $(x_n)$ has a subsequence converging to some $y$; in particular this subsequence is a Cauchy sequence. For all $\varepsilon>0$, there exist two integers $n<m$ such that $d(x_n,x_m)\leq \varepsilon$ hence $d(x,f(X)) \leq d(x_0,x_{m-n})=d(x_n,x_m) \leq \varepsilon$. Hence $d(x,f(X))=0$ i.e. $x \in f(X)$.

Glinka
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Seirios
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  • Yes, that's the idea of proof at http://math.stackexchange.com/questions/170989/a-isometric-map-in-metric-space-is-surjective . – user87690 Aug 27 '13 at 09:49
  • your claim that d(x,f(X))=0 ie. x∈f(X) is incorrect. Consider the set A={1/n:n is a positive integer}. let x=0. Then d(x,A)=0 but x is not in A. – RagingBull Mar 27 '15 at 14:44
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    @BasantSharma: The property $d(x,A)=0 \Rightarrow x \in A$ is true whenever $A$ is closed. In my answer, because $X$ is compact, $f(X)$ is indeed closed. – Seirios Mar 27 '15 at 15:37
  • yes you are right....sorry for the hurried response – RagingBull Mar 27 '15 at 19:49
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Another proof: Let $f: X \to X$ be a continuous function on (non-empty) compact and sequentialy compact topological space. Then we can define $X_0 = X$, $X_{n + 1} = f[X_n]$, $X' = \bigcap_{n < ω} X_n$. $X'$ is non-empty compact space, since it's an intersection of decreasing sequence of compacts. By sequential compactness $f: X' \to X'$ is onto. (For $x ∈ X'$ we can choose $x_n ∈ X_n$ such that $f(x_n) = x$, there is convergent subsequence $x_{n_k} \to y ∈ X'$ so $f(x_{n_k}) = x \to f(y)$ so $x = f(y)$.) Also for any $x ∈ X$ we can define $x_0 = x$, $x_{n + 1} = x_n$. Then $x_n \to X'$ in the sense that for any open $U ⊇ X'$, $x_n$ is eventually in $U$. (This is by the fact, that $x_n$ is eventually in any $X_n$ and by compactness.)

In our case ($X = M$) we have that $f: X' \to X'$ is homeomorphism and for any $x ∈ X$ $d(x_n, X') \to 0$ but also $d(f(x), X') = d(f(x), f[X']) = d(x, X')$ so $d(x_n, X')$ is constant and hence zero, so $x ∈ X'$.

user87690
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