Let $(X,d)$ be a compact metric space. Let $f:X \to X$ be an isometric embedding, which means that for each $x,y\in X$, we have: $$ d(f (x),f (y)) = d (x,y). $$
Is then $f$ automatically surjective? In other words, is it true that $$ \inf_{y\in X} d (x,f (y)) = 0 $$ for every $x\in X $?
What I've figured out so far: this is clearly true if $X $ is finite, and clearly false if $X $ is not compact.
Consider $x\notin \operatorname{im}f$. Consider the sequence $\{f^n(x)\}_{n\in\Bbb N}$, which has a convergent subsequence $\{f^{n_k}(x)\}_{k\in\Bbb N}$. Since it is convergent, you know that $$\lim_{k\to\infty}d(f^{n_{k+1}}(x),f^{n_k}(x))=0$$ Since $f$ is an isometry, this means that $$\lim_{k\to\infty}d(f^{n_{k+1}}(x),f^{n_k}(x))=\lim_{k\to\infty} d(f^{n_{k+1}-n_k}(x),x)=0$$ Which proves that indeed $x\in\operatorname{cl}(\operatorname{im}f)$.
