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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 $A \subset f(A)$?

egreg
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ART
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1 Answers1

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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.

user2566092
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  • Sorry Im writing in ipad...so. Would it last to define the notion of volume $V_{\epsilon} (A) = {x\in A | \parallel f(x)-f(y)\parallel = \epsilon \gt 0, $\epsilon$ \in \Re positive}$

    Because it is compact, in particular it is bounded,further more \epsilon is finite.

    If $f(A)=A$ you can find a Ball $B_{\delta} (y) \subset A$ such that $f(y) \notin B_{\delta} (y)$ then it happens that $\parallel f(x)-f(y)\parallel$ $\ne$ $\parallel x-y \parallel$

    $\therefore $ it is not a isometry... Contradiction

     – ART Oct 07 '15 at 20:45