Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 9 down vote accepted

That's a nice idea for a proof. I think perhaps it works well to turn it inside out, so to speak:

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.