Let \(X\) be a compact metric space and let \(f:X\to X\) be an isometry. Show that \(f\) is surjective.
Hint: for a point \(x\), find \(m<n\) such that \(f^n(x)\) is close to \(f^m(x)\).
Solution
1. Suppose, for contradiction, that \(f\) is not surjective. Then there are\(x\in X\) and \(\varepsilon>0\) such that\[ B(x,\varepsilon)\cap f(X)=\varnothing. \]
If \(f\) is not surjective, choose \[ x\in X\setminus f(X). \] An isometry is continuous, so \(f(X)\) is compact because \(X\) is compact. Hence the continuous function \[ y\longmapsto d(x,y) \] attains a minimum on \(f(X)\). Since \(x\notin f(X)\), this minimum is some positive number \(\delta\). Taking, for example, \[ \varepsilon=\frac{\delta}{2} \] gives the stated disjoint ball.
2. There exist integers \(0\leq j<k\) such that\[ d\bigl(f^j(x),f^k(x)\bigr)<\varepsilon. \]
The sequence \[ x,f(x),f^2(x),\ldots \] lies in the compact metric space \(X\), so it has a convergent subsequence. Every convergent sequence is Cauchy. Hence two distinct terms of that subsequence, say with indices \(j<k\), have distance less than \(\varepsilon\).
3. One has\[ d\bigl(x,f^{k-j}(x)\bigr)<\varepsilon. \]
Every iterate \(f^j\) is again an isometry. Therefore \[ d\bigl(f^j(x),f^k(x)\bigr) = d\bigl(f^j(x),f^j(f^{k-j}(x))\bigr) = d\bigl(x,f^{k-j}(x)\bigr). \] Step 2 gives the strict inequality.
4. The assumption that \(f\) is not surjective is impossible.
Since \(k-j\geq1\), \[ f^{k-j}(x) = f\bigl(f^{k-j-1}(x)\bigr) \in f(X). \] By step 3 this point also lies in \(B(x,\varepsilon)\), contradicting step 1.
5. The map \(f\) is surjective.
Step 4 rules out failure of surjectivity.
6. Q.E.D.
Step 5 is the required conclusion.