Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I am not sure if this statement is accurate but I feel I read that you can establish that there can be no homeomorphism from $\mathbb{R}^2 \to \mathbb{R}^3$ using jordan curve theorem. can you sketch or post a link to it.

share|cite|improve this question
up vote 4 down vote accepted

Suppose we had some homeomorphism $f:\mathbb R^2\to \mathbb R^3$. Then if $g:[0,1]\to \mathbb R^3$ were a simple closed curve, we would have that $f^{-1}(\mathbb R^3\setminus \mathrm{img}(g))=\mathbb R^2\setminus \mathrm{img}(f^{-1}\circ g)$ which is disconnected by the Jordan curve theorem. Thus since $f$ is a homeomorphism, $f(\mathbb R^2\setminus \mathrm{img}(f^{-1}\circ g))=\mathbb R^3\setminus \mathrm{img}(g)$ would be disconnected. Take any simple closed curve in $\mathbb R^3$ to get a contradiction.

share|cite|improve this answer
A late question, but is not $\mathbb{R}^3 img(g)$ disconnected? Do not all simple closed curves disconnect a sphere? – user45099 Mar 7 '13 at 18:56
@user1709828 $\mathbb R^3\setminus\mathrm{img}(g)$ is certainly not disconnected. It's 3-dimensional Euclidean space minus a simple closed curve. For some simple closed curves this may be hard to prove, but it suffices to show it for any curve, for example the unit circle in the $xy$-plane. – Alex Becker Mar 7 '13 at 21:13

Your Answer


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