# Proof that one-point compactification of $\mathbb{R}^n$ is homeomorphic with $S^n$

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?

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}

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