is an isometry injective or bijective?
An isometry is a distance preserving map. This means $f: X \to Y$ is an isometry iff $$d(x,y) = d'(f(x),f(y))$$ where $d$ and $d'$ are the respective metrics on $X$ and $Y$. An isometry is automatically injective and continuous. One need only let $\delta = \epsilon$ in the $\epsilon, \delta$ characterization of continuity) and if $f(x) = f(y)$ , then $d(f(x),f(y)) = d(x,y) = 0$ which implies $x =y$
There do exist maps which are injective and continous which are not isometries ofcourse. Something like $f: \mathbb{R} \to \mathbb{R}$ like $f(x) = x^3$ should work.
excalibirr
College Student studying mathematics. Still just a beginner but eager to learn. I really love the mathstackexchange website and am eager to work my way up through the ranks both here and in real life mathematics. I also enjoy many other things in life too. A good book , a good puzzle, a good adventure and good friends.
Updated on October 23, 2020Comments
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excalibirr about 3 years
On wolfram Alpha it defines an isometry as a bijective map between two metric spaces which preserves distance. However in my college notes it defines it as being one-one and continuous. which is the correct definition ? one to one or bijective ?
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excalibirr about 6 yearsbut does it also have to be bijective as stated in wolfram alpha, or does it just need to be injective ?
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Dionel Jaime about 6 yearsSome books/sources just add bijective in the definition and some don't.