# One point compactification of $\mathbb{R}^{n}$ is homeomorphic to $S^{n}$

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.

• 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? – Brian M. Scott Aug 22 '20 at 16:47
• @BrianM.Scott I don't see the homeomorphism given "by definition", is $\phi$ Seng described ? – jacopoburelli Aug 22 '20 at 17:04
• 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 $x_0$.) – Brian M. Scott Aug 22 '20 at 17:07
• Okay thanks, got the point. – jacopoburelli Aug 22 '20 at 17:11
• You’re welcome. – Brian M. Scott Aug 22 '20 at 17:13

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.