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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?

zyx
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user2566092
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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
  • Beaten by Harald. – Daniel Fischer Sep 16 '13 at 19:12
  • @DanielFischer Indeed it looks that way. Thanks for being willing to take the time though Daniel. :) I see you on here a lot and I appreciate the help you're willing to give. – user2566092 Sep 16 '13 at 19:14
  • No problem. Fortunately, I wasn't very far with typing when Harald posted ;) – Daniel Fischer Sep 16 '13 at 19:15
  • @zyx Thanks for the tip, I normally assume questions on here go forgotten after a couple hours, but I'll wait to accept the leading answer for a bit based on your recommendation that an even better answer matching/generalizing my intuition may come along. – user2566092 Sep 16 '13 at 19:28
  • @DanielFischer sorry to be stepping on your toes, if ever so slightly. I just couldn't help it. 8-) – Harald Hanche-Olsen Sep 16 '13 at 19:31
  • @HaraldHanche-Olsen Some time, some question, when you least expect it, I will post before you have finished typing. – Daniel Fischer Sep 16 '13 at 19:33
  • @DanielFischer Yeah, it happens to me all the time. But this time, I knew you were on the ball. You told me so yourself. And yet, and yet … I pounced. – Harald Hanche-Olsen Sep 16 '13 at 19:36

1 Answers1

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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.

Harald Hanche-Olsen
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  • 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
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    This can be seen as a reformulation of the "official proof". In your formulation, we suppose $E$ to be maximal, we prove that it must be finite (the lemma), and we obtain a contradiction by adding one more element. In the "official proof", we start with $E={x}$ and at each step, we add an element (which is $f^n(x)$). We get an infinite set, but it must be finite by your lemma. – Dabouliplop May 07 '18 at 20:29
  • @Idéophage Thanks for the insight. I'm afraid it's been so long since I saw the “official proof”, I had completely forgotten it. – Harald Hanche-Olsen May 08 '18 at 10:28
  • The "official proof" is the following. Suppose $x$ is not in the closure of ${f^n(x)}{n∈ℕ}$. Then, there is a radius $>0$ such that $∀n≥1 : d(f^n(x),x) ≥ $. Because $f$ is an isometry, this implies $∀n,m ∈ ℕ : d(f^n(x),f^m(x)) ≥ $. So, the infinite set ${f^n(x)}{n∈ℕ}$ contradicts your lemma. In the usual proof, your lemma is proved by saying that an infinite set must have an accumulation point. We take two points close enough to that accumulation point to get a contradiction. – Dabouliplop May 09 '18 at 19:11
  • In your formulation, you say "let $E$ be maximal" and get a contradiction by constructing a larger $E$. We can think of the argument as saying : we start with $E={x}$. Ah, but it is not maximal, because we can take $E={x,f(x)}$. But this is still not maximal, we can take $E={x,f(x),f^2(x)}$. Etc. We get $$-separated sets of each finite cardinal, so $X$ cannot be compact ("infinite volume"). I bet I have already seen that "reformulation pattern". – Dabouliplop May 09 '18 at 19:11