TITLE: Isometry and compactness.
QUESTION [5 upvotes]: Let $K\subset \mathbb{R}^{n}$ be a compact non-empty set and $f: K \to K$ such that $|f(x) - f(y)| \geq |x-y|$, for all $x,y \in K$. Consider the sequnce $\{f_{n}\}$ given by $f_{n} = f^{n} = f\circ \cdots \circ f$. For $a,b \in K$, let $a_{n} = f_{n}(a)$ and $b_{n} = f_{n}(b)$.
(a) Show that for any $\epsilon  > 0$, there is $k \in \mathbb{N}$ such that
  $$|a-a_{k}| < \epsilon, |b-b_{k}|<\epsilon.$$
(b) Show that $f(K)$ is dense in $K$.
(c) Show that $f$ is an isometry

My attempt
(a) I dont have any idea
(b) Given $x \in K$, take $x_{n} = f_{n}(x)$. By (a), for any $\epsilon > 0$, there is $k \in \mathbb{N}$ such that $|x - x_{k}| < \epsilon$. The sequence $\{x_{k}\}$ satisfying (a) has a convergent subsequence, since $K$ is compact. But this sequence converges to $x$.
(c) My idea was suppose that $|a - b| > |f(a) - f(b)|$ for some $a,b$ using (b), but I cannot conclude anything.

Can someone help me? Also, I'm not sure about (b). This not seems a hard problem, but I'm stuck on it.

REPLY [2 votes]: Let $\epsilon>0$. Cover $K$ with finitely many open balls of radius $\epsilon/2$, as we may, since $K$ is compact. The sequence $(a_n)$ must visit at least one such ball, say $B$, infinitely many times. Therefore there is a subsequence $(a_{n_k})$ of $(a_n)$ that lives wholly in $B$. In particular, $|a_{n_k} - a_{n_{k+1}}| < \epsilon$ for all $k$, where we also used the triangle inequality. By the expansive property of $f$, and by induction, it follows that $|a - a_{n_k - n_{k+1}}| < \epsilon$ for all $k$, that is, $(\alpha_k) = (a_{n_k - n_{k+1}})$ is a subsequence of $(a_n)$ that lives wholly in the $\epsilon$-ball centred on $a$. Taking the corresponding subsequence of $(b_n)$, and applying the same argument using compactness of $K$ and expansiveness of $f$, we may pass to further corresponding subsequences of $(a_n)$ and $(b_n)$ that live wholly in the $\epsilon$-balls centred on $a$ and on $b$, respectively. In particular, there is an index $k$ such that $a_k$ and $b_k$ have the property required in part a) of the question.
For part b), let $c \in K$. We already showed in part a) that points in the sequence $f^n(c)$ may be found that are arbitrarily close to $c$. It follows that the range of $f$ is dense in $K$, as required.
For part c), let $a, b \in K$ and suppose, for a contradiction, that $|f(a) - f(b)| = |a - b| + \eta$, where $\eta > 0$. Then for all $k$, $|a_k - b_k| \geq |a - b| + \eta$. By a suitable choice of $\epsilon$, and by applying the triangle inequality, we can contradict the assertion in part a). Conclude that $|f(a) - f(b)| = |a - b|$.