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

Let $\phi : \mathbb{R}^2\rightarrow\mathbb{R}^2$ be an isometry. Suppose $\phi$ is not surjective, that is there exists some $v \in \mathbb{R}^2$ whose fiber $\phi^{-1}(v)$ is empty. Then by the pigeonhole principle there exist $u, u' \in \mathbb{R}^2$ where $u\neq u'$ which map to the same element $\phi(u)$. But then $\phi$ is not an isometry since $d(u,u') > d(\phi(u),\phi(u))=0$.

My issue is with using pigeonhole principle for uncountable sets, which feels flawed to me.

share|cite|improve this question
up vote 6 down vote accepted

Indeed, your problem is using the pigeonhole principle for infinite sets (not even uncountable). To wit, consider the map $f: \mathbb{Z} \to \mathbb{Z}$ defined by $f(x) = 2x$. Since $f^{-1}(1)$ is empty, by your argument, $f$ must not be injective, which is clearly false.

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.