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?