# Show that using Jordan curve theorem $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^3$

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.

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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.
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