# One point compactification of $\mathbb{R}^n$ proof

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!

• Did you click the link to read about (and see the picture of) stereographic projection? Commented Aug 21, 2020 at 12:50
• The word "complements" is missing in that paragraph. $U_{\infty} \setminus \{\infty\}$ corresponds to the complement of a closed and bounded subset of $\mathbb{R}^n$. Commented Aug 21, 2020 at 12:57

Everybody should have trouble because it is wrong that

under stereographic projection the open subspaces $$U_\infty \setminus \{\infty\} \subset S^n \setminus \{\infty\}$$ are identified precisely with the closed and bounded subsets of $$\mathbb{R}^n$$.

As Daniel Fischer comments, the correct statement is

under stereographic projection the open subspaces $$U_\infty \setminus \{\infty\} \subset S^n \setminus \{\infty\}$$ are identified precisely with the complements of the closed and bounded subsets of $$\mathbb{R}^n$$.

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:

Under any homeomorphism $$h : S^n \setminus \{\infty\} \to \mathbb{R}^n$$ the open subspaces $$U_\infty \setminus \{\infty\} \subset S^n \setminus \{\infty\}$$ are identified precisely with the complements of the compact subsets of $$\mathbb{R}^n$$.

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 $$h(U_\infty \setminus \{\infty\}) = \mathbb R^n \setminus K$$.