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I'd like to understand the first proof given to me of the fact that the one point compactification of $\mathbb{R}^{n}$ is homeomorphic to $\mathbb{S}^{n}$.

The proof goes as follows : there is an initial remark about $i: \mathbb{R}^{n} \longrightarrow \mathbb{S}^{n}$ being an open embedding (with the identification of $\mathbb{R}^{n}$ as $\mathbb{S}^{n}-\left\lbrace x_{0}\right\rbrace$) and then it states that we only have to proof that the euclidean topology of $\mathbb{S}^{n}$ coincides with the Alexandrov topology on the compactification of $\mathbb{R}^{n}$.

I don't understand how checking the open subset's conditions is sufficient to deduces the homomorphism. Are we using some uniqueness of the Alexandrov topology ?

However I know there is a much simpler way, which is to proof in general that if a topological space $X$ is compact and Hausdorff then it is homeomorphic to the one point compactification of $X$ minus a point, but I'm interested in understanding this one.

Any help or hint would be appreciated.

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    $\begingroup$ I don’t understand your question: if the Euclidean topology on $\Bbb S^n$ coincides with the topology of the one-point compactification of $\Bbb R^n$ when $x_0$ is taken as the point at infinity, then by definition $\Bbb S^n$ is homeomorphic to the one-point compactification of $\Bbb R^n$. What is the problem? $\endgroup$ – Brian M. Scott Aug 22 '20 at 16:47
  • $\begingroup$ @BrianM.Scott I don't see the homeomorphism given "by definition", is $\phi$ Seng described ? $\endgroup$ – jacopoburelli Aug 22 '20 at 17:04
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    $\begingroup$ You don’t need an actual homeomorphism if you’ve proved that the topologies are the same. (Though doing that does implicitly construct a homeomorphism that is $i$ on $\Bbb R^n$ and sends $\infty$ to $x_0$.) $\endgroup$ – Brian M. Scott Aug 22 '20 at 17:07
  • $\begingroup$ Okay thanks, got the point. $\endgroup$ – jacopoburelli Aug 22 '20 at 17:11
  • $\begingroup$ You’re welcome. $\endgroup$ – Brian M. Scott Aug 22 '20 at 17:13
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Let $K$ be a compact Hausdorff space, $a\in K$, $K'=K\setminus\{a\}$ and $K'^+=K'\cup\{\infty\}$ be the one-point compactification of $K'$. Then $\phi:K'^+\to K$ given by inclusion on $K'$ and $\phi(\infty)=a$ is a homeomorphism.

One just has to prove that $\phi$ is continuous, since then $\phi$ is a continuous bijection from a compact space to a Hausdorff space, it must be a homeomorphism.

Continuity of $\phi$ at all points of $K'$ is obvious. What about continuity at $\infty$? If $U$ is an open neighbourhood of $a$ in $K$ then $\phi^{-1}(U)=(U\setminus\{a\})\cup\{\infty\}$. The complement of $\phi^{-1}(U)$ is $K\setminus U$ which is a compact subset of $K\setminus\{a\}$, so $\phi^{-1}(U)$ is open by the definition of the topology on the one-point compactification.

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  • $\begingroup$ Isn’t this simply the general result that the OP mentions at the end of the question? (Also proved here and here.) If so, I’m not sure that it really answers the question. $\endgroup$ – Brian M. Scott Aug 22 '20 at 16:52

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