Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We defined an isometry to be a bijection $f:X\rightarrow X'$ such that $d'(f(x_1),f(x_2))=d(x_1,x_2)$ $\forall x_1,x_2\in X$. Show that any isometry is a homeomorphism.

So my definition of homeomorphism is that a function $f:X\rightarrow X'$ is a homeomorphism if $f$ is a bijection and $f^{-1}$ is continuous. So I have to show that

(a) $f$ is continuous.

$\forall\epsilon>0$ pick $\delta=f^{-1}(\epsilon)$. Then it follows that $d(x_1,x_2)<\delta\implies d'(f(x_1),f(x_2))<\epsilon.$

(b) $f^{-1}$ is continuous. Is this just a reverse of (a)?

share|cite|improve this question
$\delta=f^{-1}(\epsilon)$ makes no sense. It seems you are making things way too complicated. – Harald Hanche-Olsen Mar 15 '12 at 16:26
That is not the definition of "homeomorphism". A homeomorphism is a continuous bijection whose inverse is continuous. – Chris Eagle Mar 15 '12 at 17:30

$f^{-1}(\epsilon)$ does not make sense: $f^{-1}$ is a function that maps from $X$ to $X$, not from $\mathbb{R}$ to $\mathbb{R}$. So you certainly cannot pick $\delta=f^{-1}(\epsilon)$.

To show that $f$ is continuous, note that given $\epsilon\gt 0$ if $d(x_1,x_2)\lt\epsilon$ then $d'(f(x_1),f(x_2))=d(x_1,x_2) \lt \epsilon$; this proves that $f$ is (uniformly) continuous (with $\delta=\epsilon$).

To show that $f^{-1}$ is continuous, simply note that it is an isometry, so by the first part, it is continuous as well.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.