Isometry in a compact space is surjective

Question

Let's say we have a compact metric space $(X,d)$ and a function $f: X\to X$ satisfying

$d(f(x),f(y)) = d(x,y)$, $x,y\in X$.

How can we show that this function is sujective in a simple way.

Answer 1

Let's suppose there is some element $y \in X - f(X)$, then since $X$ is a compact metric space, there exists a disjoint $\epsilon-$ neighborhood around $y$ that doesn't intersect $f(X)$ (here we've used that $f(X)$ is the continuous image of a compact space).

We're now ready to construct a sequence that has no convergent subsequence (contradicting that $X$ is compact)

We let $x_0 = y$ and then let $x_{n+1} =f(x_{n})$. We first see that $d(x_0, x_i) > \epsilon$ for all $i >0$ since $x_i \in f(X)$. Also, by our construction, for all $m > n >0$ we have that $$ d(x_m,x_n) =d(f(x_{m-1}),f(x_{n-1}) = d(x_{m-1},x_{n-1}) \cdots = d(x_{m-n},x_0) > \epsilon$$ where the first equality is from the definition of the sequence, the second equality is from the fact that $f$ is an isometry, and then we just continue in this pattern to get the final inequality. Thus, this sequence can have no convergent subsequence, contradicting that $X$ is compact. Thus $f$ must be surjective.

Answer 2

If $\{x_i\}$ is a $2\delta$-maximal packing, i.e. $d(x_i,x_j)>2\delta$, then $f(\{x_i\})$ is $2\delta$-maximal packing.

If $d(x_i,x)>2\delta $, then $\{ x_i\}$ is not maximal. Hence $\{x_i\}$ is $2\delta$-net, i.e. for any $x\in X$, there is $x_i$ s.t. $d(x_i,x)\leq 2\delta$.

Hence for $x\in X$, we have $d(f(x_i),x)\leq 2\delta$. That is, $x$ is a limit point of $f(X)$. Since $X$ is compact, then $x\in f(X)$.