Let $A\subset \mathbb{R}^{n}$ a compact subset and $f:A\rightarrow A$ an isometry (i.e , a function such that $\| f(x)-f(y)\| = \| x-y\|$ for all $ x,y \in A$).
Show that $f(A)=A$
Now, I've already shown that $f(A)\subset A$ like this:
$\forall x \in A$, $f(x) \in A$; then $f(A) \subset A$
Any hint to show that $A \subset f(A)$?
Hint: Define a notion of "volume" $V_\epsilon (A)$ which is the largest number of points you can choose from $A$ such that all points are at least $\epsilon$ apart. Note by compactness, this is a finite number. If $f(A) \neq A$, then you can find a ball in $A$ that does not contain any points in $f(A)$. Use this (and the isometry) to contradict the notion of volume.