$\mathbb{R}^n$ not homeomorphic to open subset of $\mathbb{R}^m$ I am a little confused on this topic, I have managed to prove that if $n < m$ then this is true, but the proof does not seem to work if $m < n$. Indeed, if $n < m$ and $f :\mathbb{R}^n \to U$ is a homeomorphism  with $U$ being an open subset of $\mathbb{R}^m$, then $\mathbb{R}^n$ will also be homeomorphic to the subset of $\mathbb{R}^m$ consisting of tuples where the last $m-n$ entries are zero, this set $A$ is closed, and so we have an homeomorphism $F: U \to A$, but since $U$ is open so must $A$ be by the invariance of domain theorem, but $A$ can't be both close and open since the only such sets in $\mathbb{R}^m$ is the empty set and $\mathbb{R}^m$ itself, contradiction.
But what if $m < n$? This proof does not seem to work then?
 A: Instead of considering a homeomorphism $\mathbb{R}^n\to U$ for open $U\subseteq\mathbb{R}^m$ consider a homeomorphism $f:V\to U$ for $V\subseteq\mathbb{R}^n$ and $U\subseteq\mathbb{R}^m$ both open. This case can be proved by modyfing a bit your proof:
Assume that $n<m$ and let $B\subseteq V$ be an open ball. We have now:
$$\phi:\mathbb{R}^n\to\mathbb{R}^m\mbox{ = the embedding onto first n coordinates}$$
$$g:B\to \mathbb{R}^n\mbox{ = a homeomorphism}$$
$$h:B\to f(B),\ h(x)=f(x)$$
Obviously $h$ is a homeomorphism. Note that $f(B)$ is open in $U$ (and hence in $\mathbb{R}^m$) because $f$ is a homeomorphism. Thus we have a well defined function:
$$\tau:f(B)\to\mathbb{R}^m$$
$$\tau=\phi\circ g\circ h^{-1}$$
Note that $\tau$ is an injective continuous function and $\mbox{im}(\tau)=\mbox{im}(\phi)$ so by the invariance of domain $\mbox{im}(\phi)$ has to be open but by the definition of $\phi$ it is closed. Since $\mathbb{R}^m$ is connected then $\mbox{im}(\phi)$ has to be either whole $\mathbb{R}^m$ or empty. But it is neither. Contradiction. $\Box$
Now the case $m<n$ is solved automatically by considering $f^{-1}$.
