# 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?

• 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. 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. 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 ;) 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. Sep 16 '13 at 19:03
• @HagenvonEitzen yes you're right, I had to use compactness to deduce that there was $\delta > 0$. Sep 16 '13 at 19:09

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.

• 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. 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
• 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. May 7 '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. May 8 '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 $𝛿>0$ such that $∀n≥1 : d(f^n(x),x) ≥ 𝛿$. Because $f$ is an isometry, this implies $∀n,m ∈ ℕ : d(f^n(x),f^m(x)) ≥ 𝛿$. 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. May 9 '18 at 19:11