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 :(
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1$\begingroup$ 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 :) $\endgroup$– rschwiebNov 15, 2013 at 17:47
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$\begingroup$ I completely understand. This is my question, and all that was given. And I dont even know where to start? $\endgroup$– R.ANov 15, 2013 at 17:49
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2$\begingroup$ 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 :) $\endgroup$– rschwiebNov 15, 2013 at 17:51
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$\begingroup$ Cheers, I have been doing so much topolgy. Last minute revision doesn't help. And nothing is going in lol. Thanks:) $\endgroup$– R.ANov 15, 2013 at 17:54
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3 Answers
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Clue: any isometry preserve distance and the distance between two points is zero if and only if they are the same point.
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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.
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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.
