I was recently reading about one-point compactification, and a proof of how the one-point compactification of $\mathbb{R}^n$ is homeomorphic to $\mathbb{S}^n$.The proof is example 4.1 at https://ncatlab.org/nlab/show/one-point+compactification , which I have attached here as an image:
I was having trouble understanding the identifications mentioned in the second last paragraph- if anybody could help provide some intuition behind how this can be seen, it would be greatly appreciated!
Best Answer
Everybody should have trouble because it is wrong that
As Daniel Fischer comments, the correct statement is
It is well-known that the closed and bounded subsets of $\mathbb{R}^n$ are precisely the compact subsets of $\mathbb{R}^n$. So let us look at the following generalization:
If you want, you can take $h$ = stereographic projection, but it is irrelevant.
In fact, the set $C = (S^n \setminus \{\infty\}) \setminus (U_\infty \setminus \{\infty\}) = S^n \setminus U_\infty$ is a closed subset of $S^n$, hence it is compact because $S^n$ is compact. Since $C \subset S^n \setminus \{\infty\}$, its image $h(C) \subset \mathbb{R}^n$ is compact. But now $h(U_\infty \setminus \{\infty\}) = h((S^n \setminus \{\infty\}) \setminus C) = h(S^n \setminus \{\infty\}) \setminus h(C) = \mathbb{R}^n \setminus h(C)$.
Conversely, if $K \subset \mathbb{R}^n$ is compact, then $h^{-1}(K) \subset S^n \setminus \{\infty\}$ is compact, thus it is closed in $S^n$ and $U_\infty = S^n \setminus h^{-1}(K)$ is an open neighborhood of $\infty$ in $S^n$. Clearly $h(U_\infty \setminus \{\infty\}) = \mathbb R^n \setminus K$.