# Distance preserving map on finite dimensional normed space E is a bijection

The problem is to show that for $E$ a finite dimensional normed space and $T:E\to E$ a map (not necessarilly linear) such that $\Vert T(x)-T(y)\Vert =\Vert x-y\Vert \forall x,y\in E$is a bijection.

1-1 is trivial. Onto I'm trying to use the fact that either the sphere or ball is compact to get it, but I'm hitting a wall. My efforts:

Since $T$ is Lipschitz, it's continuous. Let $T(0)=x_0$ . Then we have $\Vert T(x)-x_0 \Vert=\Vert x \Vert \forall x\in E$. Look at the unit closed ball (or sphere, not sure which is easier) around $0$. (If it works for the unit ball, it will work equally for any other distance, and thus replacing 1 with every other number will get me the whole space). Since E is finite dimensional, it is compact. Thus $T(B(0,1))$ is compact. We have $\forall x\in B(0,1),\Vert x \vert \le 1$, hence $\Vert T(x) -x_0 \Vert \le 1$, thus $T(x)\in B(x_0,1)$, so $T(B(0,1))\subseteq B(x_0,1)$.

I want equality, because then I'd be done, but I can't seem to get it, even though it's obvious that since it preserves distance it must be equal. I know it's a compact subset of a compact set, so if I had a point in one and not the other then the distance to the closed set would be positive...trying to construct a sequence then use sequential compactness hasn't gotten me anywhere.

Am I barking up the wrong tree? My other thought seemed highly silly, though I did try it for a bit....induction on the dimension of the space. That got me that every point with a 0 in one of the components is hit by $T$, but I couldn't get past that.

• See the proof here – Omnomnomnom Jan 29 '15 at 0:58
• @Omnomnomnom Hmm, seems to be via the lebesgue number then? And I'm pretty sure I can't assume wlog that T(0)=0 like he is, or things would simplify immensely – Alan Jan 29 '15 at 1:07
• Sure you can: $x \mapsto T(x)$ is an isometry iff $x \mapsto T(x) - T(0)$ is an isometry. – Omnomnomnom Jan 29 '15 at 1:09
• What norm are you using? – copper.hat Jan 29 '15 at 1:46
• I realise that, but if it was the Euclidean norm, then things are a little easier. – copper.hat Jan 29 '15 at 4:46