# One-Point-Compactification

It is well known that every topological space has a compactification. We therefore consider the one-point extension $$(Y, T)$$ of a topological space $$(X, V)$$ which is a compact space with a inclusion map $$\iota$$ as the embedding. Therefore $$(\overline{\iota(X)}, T_{\iota(X)}, \iota)$$ is the compactification of $$(X, V)$$.

We also know that the one-point extension is hausdorff if and only if $$(X,V)$$ is hausdorff and locally compact.

Is it true that for a compact hausdorff space $$(Y, T)$$ we can construct a topological space $$(X, V)$$ and an embedding $$\iota$$ such that $$(Y, T, \iota)$$ is a compactification of $$(X,V)$$ and $$(Y,T)$$ is a one-point extension of $$(X,V)$$ up to homoemorphism? So is it possible to somehow reverse the construction of a one-point extension of a topological space $$(X, V)$$?

I thought that we can reverse the construction of a one-point extension by taking a limit point $$x \in Y$$ of $$Y$$ and look at $$X := Y \setminus \lbrace x \rbrace$$ where $$Y = \lbrace x \rbrace \cup Y \setminus \lbrace x \rbrace$$ and show that \begin{align} T &= \lbrace U \subseteq Y : x \not\in U, \iota^{-1}(U) \mbox{ is open in } X \rbrace \\&\cup \lbrace U \subseteq Y : x \in U, X \setminus \iota^{-1}(U) \mbox{ is compact and closed in } X \rbrace \end{align} Since we look at one point extensions up to homeomorphism we can suppress the notation where $$Y := Y \setminus \lbrace x \rbrace \sqcup \lbrace x \rbrace$$ and $$\iota: Y \setminus \lbrace x \rbrace \rightarrow Y \setminus \lbrace x \rbrace \sqcup \lbrace x \rbrace$$ where $$X \sqcup Y$$ means the union of $$\iota_X(X)$$ and $$\iota_Y(Y)$$ where $$\iota_X: x \mapsto (x,0)$$ and $$\iota_Y: y \mapsto (y,1)$$. It is clear that since $$(Y, T)$$ is a one-point extension of $$(X,V)$$ that it is unique up to homeomorphism.

Are my thoughts so far true or am I misunderstanding something?

• The Aleksandrov extension is only a compactification (in the sense that $\iota[Y]$ is dense in $X$) when $Y$ is not compact to start with, of course. I take it in the most liberal sense, so I allow compact $Y$; i.e. the Aleksandrov extension always is defined and is compact and $\iota$ is an embedding. – Henno Brandsma Apr 9 at 15:14
• If someone is interested in the proof of the construction I recently found an answer: math.stackexchange.com/a/427131/904739. – Orb Apr 9 at 15:50

## 1 Answer

It's true: for a compact Hausdorff space $$Y$$ we can prove that $$Y$$ is the Aleksandrov extension (the common name for the one-pont extension) of $$Y\setminus \{x\}$$ for any $$x \in Y$$.

Of course $$Y$$ can be the Aleksandrov extension of several non-homomorphic spaces when $$Y$$ is non-homogeneous, so it's only a one-sided inverse..