# How do I prove that an isometry is injective?

How do I prove that an isometric embedding is necessarily injective?

This is my question. I study topology. Someone help, please; I don't have a clue :(

• Hi, welcome to math.SE. We understand you may have trouble getting started, but really posters are expected to supply more than just a problem statement. Any partial progress or past failures or relevant thoughts on the problem are good to include. If you post a lot of questions like this, they're likely to be closed and you're likely to be snubbed. :S So the antidote is to bring as much to the table as you can :) Nov 15, 2013 at 17:47
• I completely understand. This is my question, and all that was given. And I dont even know where to start?
– R.A
Nov 15, 2013 at 17:49
• You start with what the words mean. If the map weren't injective, then it would move two points to a single point. But originally the points were a positive distance $\epsilon$ apart, and after the map moves them distance $0$ apart. Since isometries preserve distance, $\epsilon=0$, a contradiction. It hardly gets any more straightforward than this :) Nov 15, 2013 at 17:51
• Cheers, I have been doing so much topolgy. Last minute revision doesn't help. And nothing is going in lol. Thanks:)
– R.A
Nov 15, 2013 at 17:54

Clue: any isometry preserve distance and the distance between two points is zero if and only if they are the same point.

Assume $x\ne y$, then $$\|T(x)-T(y)\|=\|T(x-y)\|=\|x-y\|\ne 0,$$ so $T(b)\ne T(b^\prime)$.

This is another way of writing essentially the same proof as written before.

An isometry is a map $T$ such that $\lvert \lvert T(a) \rvert \rvert = \lvert \lvert a \rvert \rvert$ for any $a$ in your space.

Suppose that $T(a)=0$ then $T$ would be injective (by definition) if this implied that $a=0$.

$$0=\lvert \lvert 0 \rvert \rvert =\lvert \lvert T(a) \rvert \rvert = \lvert \lvert a \rvert \rvert$$

so by the properties of norms this obviously implies that $a=0$ and we are finished.