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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: enter image description here

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!

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

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

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