Let \((X, d)\) be a metric space. If \(f: X \to X\) satisfies the condition
\[ d(f(x), f(y)) = d(x, y) \]
for all \(x, y \in X\), then \(f\) is called an isometry of \(X\). Show that if \(f\) is an isometry and \(X\) is compact, then \(f\) is bijective and hence a homeomorphism. [Hint: If \(a \notin f(X)\), choose \(\epsilon\) so that the \(\epsilon\)-neighborhood of \(a\) is disjoint from \(f(X)\). Set \(x_1 = a\), and \(x_{n+1} = f(x_n)\) in general. Show that \(d(x_n, x_m) \geq \epsilon\) for \(n \neq m\).]
Solution
Goal: Prove that an isometry \(f: X \to X\) on a compact metric space \((X, d)\) is surjective, and deduce that \(f\) is a homeomorphism.
1. \(f\) is injective and continuous:Proof:
1.1. If \(f(x) = f(y)\), then \(d(x, y) = d(f(x), f(y)) = 0\), which implies \(x = y\). Thus \(f\) is injective.
1.2. For any \(\varepsilon > 0\), setting \(\delta = \varepsilon\) ensures \(d(x, y) < \delta \implies d(f(x), f(y)) = d(x, y) < \varepsilon\). Thus \(f\) is uniformly continuous.
2. The image \(f(X)\) is compact and closed in \(X\):Proof: Since \(X\) is compact and \(f\) is continuous (<1>1), \(f(X)\) is a compact subset of the metric space \(X\), hence \(f(X)\) is closed in \(X\).
3. \(f\) is surjective (\(f(X) = X\)):Proof by contradiction:
3.1. Suppose \(f(X) \neq X\), so there exists a point \(a \in X \setminus f(X)\).
3.2. Since \(f(X)\) is closed and \(a \notin f(X)\), the distance \(\varepsilon = d(a, f(X)) = \inf_{y \in f(X)} d(a, y)\) is strictly positive: \(\varepsilon > 0\).
3.3. Define a sequence \((x_n)_{n=1}^\infty\) in \(X\) inductively by \(x_1 = a\) and \(x_{n+1} = f(x_n) = f^n(a)\) for all \(n \ge 1\).
3.4. For any positive integers \(n > m \ge 1\), since \(f^{m-1}\) is an isometry:\[d(x_n, x_m) = d(f^{m-1}(x_{n-m+1}), f^{m-1}(x_1)) = d(x_{n-m+1}, x_1) = d(x_{n-m+1}, a).\]
3.5. Since \(n > m\), \(n - m + 1 \ge 2\), so \(x_{n-m+1} = f(x_{n-m}) \in f(X)\).
3.6. By definition of \(\varepsilon\), \(d(x_{n-m+1}, a) \ge \varepsilon > 0\).
3.7. Thus \(d(x_n, x_m) \ge \varepsilon > 0\) for all \(n \neq m\).
3.8. The sequence \((x_n)\) contains no Cauchy subsequence, so it has no convergent subsequence.
3.9. This contradicts the sequential compactness of the compact metric space \(X\).
3.10. Therefore \(f(X) = X\), so \(f\) is surjective.
4. \(f\) is a homeomorphism:Proof:
4.1. By 1 and 3, \(f\) is a continuous bijection.
4.2. A continuous bijection from a compact space to a Hausdorff space is a closed map, hence a homeomorphism (its inverse \(f^{-1}\) is continuous and is an isometry as well). Q.E.D.