# Can this intuition give a proof that an isometry $f:X \to X$ is surjective for compact metric space $X$?

A prelim problem asked to prove that if $X$ is a compact metric space, and $f:X \to X$ is an isometry (distance-preserving map) then $f$ is surjective. The official proof given used sequences/convergent subsequences and didn't appeal to my intuition. When I saw the problem, my immediate instinct was that an isometry should be "volume-preserving" as well, so the volume of $f(X)$ should be equal to the volume of $X$, which should mean surjectivity if $X$ is compact. The notion of "volume" I came up with was the minimum number of $\epsilon$-balls needed to cover $X$ for given $\epsilon > 0$. If $f$ were not surjective, then because $f$ is continuous this means there must be a point $y \in X$ and $\delta > 0$ so that the ball $B_\delta(y)$ is disjoint from $f(X)$. I wanted to choose $\epsilon$ in terms of $\delta$ and use the fact that an isometry carries $\epsilon$-balls to $\epsilon$-balls, and show that given a minimum-size cover of $X$ with $\epsilon$ balls, that a cover of $X$ with $\epsilon$-balls could be found with one fewer ball if $f(X) \cap B_\delta(y) = \emptyset$, giving a contradiction. Can someone see a way to make this intuition work?

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There's not much left. Basically, choose $\varepsilon < \delta/2$, and you have your contradiction. Of course there's a little work to do to make the notions precisely defined. – Daniel Fischer Sep 16 '13 at 18:47
My intuition exceeds my formal math abilities...can you post the filling in of the gaps as an answer? I just can't see how to make the proof go. Sorry to be a bother that I can't figure this out myself. – user2566092 Sep 16 '13 at 18:51
Will do. Need to think up a catchy name for the minimal number of balls required, however, that may take a while ;) – Daniel Fischer Sep 16 '13 at 18:53
Where do you get $\delta>0$ from? This already needs compactness: $x\mapsto x-1$ is an isometry $X\to X$ for $X=\mathbb R\setminus \mathbb N$ and leaves out just a simgle point. – Hagen von Eitzen Sep 16 '13 at 19:03
@HagenvonEitzen yes you're right, I had to use compactness to deduce that there was $\delta > 0$. – user2566092 Sep 16 '13 at 19:09

Lemma. Assuming $X$ is a compact metric space, for each $\delta>0$ there is a finite upper bound to the number of points in $X$ with a pairwise distance $\ge\delta$. (Let us call such a set of points $\delta$-separated.)
Proof. $X$ is totally bounded, so there exists a finite set $N$ of points in $X$ so that every $x\in X$ is closer than $\delta/2$ to some member of $N$. Any two points in $B_{\delta/2}(x)$ are closer together than $\delta$, so there cannot be a $\delta$-separated set with more members than $N$.
Now let $f\colon X\to X$ be an isometry and not onto. Let $x\in X\setminus f(X)$, and let $\delta>0$ be the distance from $x$ to $f(X)$. Let $E\subseteq X$ be a $\delta$-separated set with the largest possible number of members. Then $f(E)\cup\{x\}$ is such a set with more members. Contradiction.
Thanks, I was really hoping there was a way to make "isometries are volume-preserving" into a proof. Your very closely related notion of a maximal $\delta$-separated set basically also captures what I meant by volume, so this definitely satisfies the hope I had. – user2566092 Sep 16 '13 at 19:19
+1. I suspect there is a metric-free argument that works in many categories including sets, topological spaces, and metric spaces, and a related axiomatization of when this property is true for maps from $X$ to an isomorphic sub-object. – zyx Sep 16 '13 at 19:49