Let $(X,d)$ be a compact metric space, where $f: X \rightarrow X$ is a distance preserving map, ie, $\forall x, y \in X$ we have that $d(f(x),f(y)) = d(x,y)$.

a) Show that f is injective. b) Show that $\forall x \in X$ and every open ball of radius $\epsilon$ centered at $x$, $B_{\epsilon}(x)$, one of the balls in the sequence: $f(B_{\epsilon}(x)),f(f(B_{\epsilon}(x))), ...$ has non-empty intersection with $B_{\epsilon}(x)$

c) Use part b) or another reasoning to show that f is surjective.

I proved part a), but am stuck on part b). I'm thinking of assuming that all the images (composed any amount of times) of the ball all have empty intersection with the original ball, but am having trouble trying to concretize what the image of the ball should be or look like.


HINT: Let $f^{(0)}$ be the identity function on $X$, and for $n\in\Bbb N$ let $f^{(n+1)}=f\circ f^{(n)}$. Note that $f^{(1)}=f$. For $n\in\Bbb N$ let $B_n=f^{(n)}[B_\epsilon(x)]$; $\langle B_n:n\in\Bbb N\rangle$ is your sequence of balls in (b).

Suppose that there are $m,n\in\Bbb N$ such that $m<n$ and $B_m\cap B_n\ne\varnothing$, and let $y\in B_m\cap B_n$.

  • Show that there is a $z\in B_0$ such that $f^{(m)}(z)=y$.
  • Show further that $z\in B_{n-m}$.
  • Conclude that $B_0\cap B_{n-m}\ne\varnothing$.

This shows that if $B_0\cap B_n=\varnothing$ for each $n\ge 1$, then the family $\{B_n:n\in\Bbb N\}$ is pairwise disjoint.

  • Use the compactness of $X$ to show that the family $\{B_n:n\in\Bbb N\}$ cannot be pairwise disjoint.
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