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I am seeking a generalization of the statement that the one-point compactification of $\mathbb{R}$ is homeomorphic with $S^1$, more specifically,

Show the one-point compactification of $\mathbb{R^n}$ is homeomorphic with $S^n$.

Let me preface this first with the statement that I do not actually know if this proposition is true, though I think it is likely. If so, a future question could be, what about for $\mathbb{R}^{\omega}$? $\mathbb{R}^J$? I would assume not so.

This problem has already been asked/responded to for n=1 and n=2, but both of these involved an actual construction of a homeomorphism.

Here is my proof for $n=1$:

Assume the following lemma:

(1) If $X$ and $Y$ are locally compact Hausdorff spaces which are homeomorphic, then their one-point compactifications, denoted $\bar{X}$ and $\bar{Y}$, are homeomorphic.

Constructing the new homeomorphism is straightforward; simply take the homeomorphism $f:X\rightarrow Y$ and define $f':X'\rightarrow Y'$ as $f(x)$ on $X$, and $Y'-Y$ on $X'-X$.

(2) The one-point compactification of $S^{1} \setminus \{ (0,-1)\}$ is $S^1$, as can be readily checked.

Thus a proof that $\mathbb{R}$ and $S^{1} \setminus \{ (0,-1)\}$ are homeomorphic is sufficient. For ease, use polar coordinates.

(3) Clearly $f:\mathbb{R}\rightarrow (-\pi,\pi)$ defined by $2\tan^{-1}(x)$ is a homeomorphism, being order-preserving and surjective. Define an order relation on $S^1$ by its $\theta$ coordinate having the same order type as $(-\pi,\pi)$.

(4) Then define $g:(-\pi,\pi)\rightarrow S^1 \setminus \{(0,-1)\}$ by sending each point of $(-\pi,\pi)$ to the point of $S^1 \setminus \{(0,-1)\}$ with that $\theta$ coordinate. Again, $g$ is order-preserving and surjective, so $g$ is a homeomorphism.

(5) Therefore, $g\circ f:\mathbb{R}\rightarrow S^{1} \setminus \{ (0,-1)\}$ is a homeomorphism. Apply (1). $\blacksquare$

Most importantly, the steps having to do with the order relation of $\mathbb{R}$ and $S^1$ are not easily scalable. More unappealing is the other, more straightforward method of constructing an actual homeomorphism between $\mathbb{R}^n\cup \{\infty\}$ and $S^n$. How could I amend my proof to make it scalable?

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I'm just expanding on Pink Panther's comment. If I understood correctly, your idea was to

  • consider a space $Y$ such that $Y \equiv \mathbb{R}$ and its compactification is $Y^* \equiv \mathbb{S}^1$, and then
  • use that $X \equiv Y$ implies $X^* \equiv Y^*$.

Moreover, since it is clear that the compactification of $\mathbb{S}^1 \setminus \{(0,-1)\}$ is the $1$-sphere, it suffices to provide an isomorphism from this space to $\mathbb{R}$.

So, what you can do to generalize this is to find $Y_n$ such that $Y_n \equiv \mathbb{R}^n$ and $Y_n^* \equiv \mathbb{S}^n$. By the argument above, it suffices to find a homeomorphism

$$ p : \mathbb{S}^n \setminus \{N\} \to \mathbb{R}^n $$

for some point $N$ of the $n$-sphere. Here's where the stereographic projection comes into play.

Geometrically, imagine the $n$-sphere embedded into $\mathbb{R}^{n+1}$ and for convinience, pick $N = e_{n+1}$ the north pole. Then, for any other point $q$ of the sphere, the line $\vec{Nq}$ intersects the plane $\Pi := \{x_{n+1} = 0\}$ of the 'floor' exactly once. Call this point $p(q)$. Moreover, one can see that this mecanism reaches every point of $\Pi$, which can be identified with $\mathbb{R}^n$.

Concretely now, via a calculation that I ommit (but I encourage you to figure out yourself before checking the literature), we can define

$$ \begin{align} p :\ & \mathbb{S}^n \setminus \{N\} \to \mathbb{R}^n\\ &(x,t) \mapsto \frac{1}{1-t}x \end{align} $$

Note that this is well defined, since the only point in the sphere with $t = 1$ is the north pole $N = (0,\dots, 0,1)$ which we have excluded.

You can check that this is a homeomorphism with inverse

$$ \begin{align} p :\ & \mathbb{R}^n \setminus \{N\} \to \mathbb{S}^n\\ &y \mapsto \frac{1}{\|y\|^2+1}(2y,\|y\|^2-1) \end{align} $$

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    $\begingroup$ Thank you; this was clear, concise, and was exactly what I was looking for. $\endgroup$ – Laurel Turner Aug 4 '19 at 1:58

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