TITLE: Prove that $\exists k\in \mathbb{N}^*$ such that $\|a-a_k\|<\varepsilon$
QUESTION [6 upvotes]: Let $E$ be a normed linear space, $C$ compact and $f:C\to C$ a function such that $\|f(x)-f(y)\|\geq \|x-y\|$ for all $x,y\in C$. Then $f$ is an isometry.

Note: I'm having trouble trying to prove it! I feel stuck. My approach: Some hint says define: $a_0=a$, $a_n=f\circ ...\circ f(a)=f^{n}(a)$ and similar $b_n=f^{n}(b)$. Added: prove that for all $\varepsilon>0$ there is $k\in \mathbb{N}^*$ such that $\|a-a_k\|<\varepsilon$ and $\|b-b_k\|<\varepsilon$
How can I deduce from this that $\|a_1-b_1\|\leq \|a-b\|+2\varepsilon$? Because in that case I know that $\|f(a)-f(b)\|\leq \|a-b\|$ and since $a,b$ were arbitrary we are done!
Please any help in proving this two things!!
My attempt: $a_n$ have (by compacity) a subsequence which converges, say $a_{n_k}\to z\in C$. Can I say that $z=a$?? What I know is that for some $x_1,...x_p\in C$ we have that $C\subseteq \bigcup_{i=1}^p B(x_i,\varepsilon)$.
Thanks I really feel stuck!

REPLY [3 votes]: Let $a_n=f^n(a)$ then we must prove that for arbitrary $\epsilon$ there exist natural K such that |$a-a_K$|<$\epsilon$
So assume that it's not true i.e. there exist $\epsilon_0$ s.t. |$a-a_k$|$\ge\epsilon_0$ for arbitrary natural $k$
Since C is compact , there are a finite number of points $x_1 , ... , x_n$ s.t. C $\subset$ $\cup D(x_i,\frac{\epsilon_0}2)$ where $D$ means disk
Then the above statement tells us that $a$ $\in$ $D(x_l,\frac{\epsilon_0}2)$ for some $l$ and $a_k$ $\in$ $D(a_i,\frac{\epsilon_0}2)$ where $i\ne l$
So there exists some K s.t. $a_1 , a_K$ are in same disk (without loss of generality) but then it is a contradiction to
$\epsilon_0$ $\le$ |$a-a_{K-1}$| $\le$ |$a_1-a_K$| $\le$ |$a_1-x_{i_0}$|+|$x_{i_0}-a_K$| $<$ $\frac{\epsilon_0}2+\frac{\epsilon_0}2$ = $\epsilon_0$ where $i_0$ is same disk's center
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