I have a problem about a compact metric space $ \mathbf{X} $ in which there is a mapping from $ \mathbf{X}$ to $ \mathbf{x}$ and the distance function is such that $d( f(x), f(y)) = d(x, y)$.
I have proved the rest of the questions asked about it, but can't seem to find a rigorous way to prove this one point: for every point $x \in \mathbf{X} $ and every positive real number $ \epsilon > 0$, there is an $n$ such that $d(f^n(x); x) < \epsilon$ (intuitively, this means that after $n$ applications of $f$, we get back to our starting point $x$ up to a very small error).
edit1: by $f^n(x)$ i mean f(f(...f(x))..) n times.
Suppose $f^{m}(x)=f^{n}(x)$ for certains $m\neq n$ natural numbers, then without loss of generality suppose $m>n$ and you get $f^{m-n}(x)=x$, and you conclude. Otherwise $n\mapsto f^{n}(x)$ is injective, which gives you an infinite subset of a compact space: it must have an accumulation point (according to the metric topology). Pick an arbitrary $\epsilon >0$ and you have $ \exists p \in \mathbb{N}$ such that $d(f^{m}(x),f^{n}(x))< \epsilon$, basically by the triangular inequality. The thesis follows using the fact that $f$ is an isometry.