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.
Best Answer
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.