TITLE: Showing that $\Bbb R^3$ is not homeomorphic to $S^3$.
QUESTION [7 upvotes]: I want to show that $\Bbb R^3$ is not homeomorphic to $S^3$.
Now this means I want to show there is no bijective function, $f$, from $\Bbb R^3$ such that both $f$ and $f^{-1}$ are continuous.

I can't seem to solve this, however some thoughts come to mind:

If there is no bijective function I am done.
All bijective functions cannot be continuous(for some reason) and I'm done.


For the first one, this seems as though it would be true. $\Bbb S^3$ 'lives in' $\Bbb R^4$ but I suppose it is bounded, whereas $\Bbb R^3$ is a lower dimension space, but it is unbounded. Hmmm.
For the second one, I can think geometrically that we can't wrap $\Bbb R^3$ into a fourth dimensional ball, such that points that are close in $\Bbb R^3$ are still close in $S^4$, but no idea how to make this rigorous.

Essentially my problem comes down to: Showing two things are homeomorphic is logically, but showing they are not seems really hard. Thanks

REPLY [17 votes]: $S^3$ is compact, while $\mathbb{R}^3$ is not. Since any continuous function $f:S^3\rightarrow \mathbb{R}^3$ maps compact subsets of $S^3$ to compact subsets of $\mathbb{R}^3$, it can't be surjective (or else $f(S^3)=\mathbb{R}^3$ is also compact).