Is $\mathbb{E}^n$ homeomorphic to $B^n-S^{n-1}$?

Question

Where $B^n$ and $S^{n-1}$ are unit ball and sphere in n-dimensional Euclidean space. I read it in Armstrong's Basic topology,page 69,but I don't know how to prove.

Answer 1

Consider the map $$ x \mapsto \frac{1}{1-\|x\|} x $$ which is a homeomorphism from one to the other.

How'd I come up with this? I spent a moment thinking about the case $n = 1$, looking for a map from $$ -1 < x < 1 $$ to the whole real line. I suppose i could have used something like "tangent" as well, but $1/(1-|x|)$ seemed pretty simple. You could make it smoother by writing $1/(1-|x|^2)$, or go all the way to "tangent" if you wanted complete smoothness, but since all you wanted was continuity, I went with the simple one.

Answer 2

The closed ball minus its boundary sphere is just the interior of the closed ball, i.e. the open unit ball.

And this is homeomorphic to the whole space in any normed space : $h(x) = \frac{1}{1+\|x\|}x$ works as a homeomorphism from the whole space to its open unit ball.