# Alternate Stone-Čech compactification definition implies Hausdorffness of $\beta X$?

Given the following definitions:

Def (Compactification): $$(\iota, Y)$$ is a compactification of a topological space $$X$$, if $$Y$$ is compact and $$\iota : X \to Y$$ is an embedding such that $$\iota(X)$$ is dense in $$Y$$.

Def (Stone-Čech compactification): A compactification $$(\beta, \beta X)$$ of topological space $$X$$ is a Stone-Čech compactification of $$X$$, if for every compact Hausdorff space $$Y$$ and continuous function $$f : X \to Y$$ there exists a unique continuous function $$g : \beta X \to Y$$ such that $$f = \beta \circ g$$.

Often the Stone-Čech compactification is defined with $$\beta X$$ being necessarily Hausdorff. Can we deduct that $$\beta X$$ is Hausdorff from the above definition? In other words: is the above definition equivalent to the common definition?

My thoughts on this:

1. If for every distinct $$x, y \in \beta X$$ there is a continuous function $$g : \beta X \to Y$$ with $$g(x) \neq g(y)$$ and $$Y$$ being a compact Hausdorff space, then $$\beta X$$ is Hausdorff: Since $$Y$$ is Hausdorff, we can find open disjoint neighborhoods $$U, V$$ of $$g(x)$$ and $$g(y)$$, thus $$g^{-1}(U)$$ and $$g^{-1}(V)$$ are open disjoint neighborhoods of $$x$$ and $$y$$.

2. $$X$$ must be completely regular for the Stone-Čech compactification to exists. Can we use this to show existence of $$g$$ with $$g(x) \neq g(y)$$ in (1)?

3. Kown constructions of the Stone-Čech compactification turn out to be Hausdorff. We can show this is unique up to homeomorphism via the universal property of the Stone-Čech compactification, but only if all other possible Stone-Čech compactifications are necessarily Hausdorff as well.

• Compactifications are normally defined to be Hausdorff compactifications. If $X$ is a $T_1$ space, then its Wallman compactiification is a non-Hausdorff compactification with the required extension property. Commented Feb 23, 2022 at 12:45
• You have to add Hausdorff compactification to def. 2 in order to really get the definition of Stone-Čech compactification. Just compactification is not enough, see my answer. Commented Feb 23, 2022 at 13:47

The short answer is no, we cannot deduce $$T_2$$ from the above definition. There is a classic construction, the Wallman compactification $$\omega X$$, based on the lattice of closed sets of $$X$$, which is defined for $$X$$ that are $$T_1$$. $$\omega X$$ is compact (not necessarily Hausdorff), $$X$$ embeds densely into it and every continuous map from $$X$$ to a compact Hausdorff space $$Z$$ can be extended to $$\omega X$$. For full details and construction, see
Engelking's book General Topology (revised edition, 1989). It's also known that $$\omega X \text{ is } T_2 \iff X \text {is } T_4$$ (so normal and $$T_1$$), and in that latter case $$\omega X$$ and the classically constructed $$\beta X$$ (as a Hausdorff compactification for any Tychonoff space $$X$$) are equivalent (and in particular homeomorphic).

So $$\omega X$$ obeys 2 (and 1), while not always being $$T_2$$ (take your favourite Tychonoff non-normal space as $$X$$ etc.)

As to your thoughts at the end, yes, on $$\beta X$$, like on any Tychonoff (includes $$T_1$$) space we can separate points by closed functions, yes. But if $$X$$ is Tychonoff, we have a function $$f: X \to [0,1]$$ separating $$x_1 \neq x_2$$ in $$X$$ in the sense that $$f(x_1)=0$$ and $$f(x_2)=1$$ and we can extend it to $$\omega X$$. But that doesn't help to prove Hausdorffness of $$\omega X$$! For that we have to separate the extra points in $$\omega X\setminus X$$ as well.. The remarks above show that we cannot expect anything from this route in general.

A really easy way to see this doesn't work is to consider what happens if $$X$$ is already compact. Then $$(1_X,X)$$ trivially satisfies your definition of a Stone-Cech compactification of $$X$$, but $$X$$ may not be Hausdorff.

Another counter-example, in which $$X$$ is a $$T_{3\frac 1 2}$$ space: Let $$p=\langle \omega, \omega_1\rangle$$ and $$X^*=(\omega +1)\times (\omega_1+1)$$ and $$X=X^*\setminus \{p\},$$ where $$\omega +1$$ and $$\omega_1 +1$$ each have the $$\in$$-order topology.

Then id$$_X:X\to X^*$$ is the Cech-Stone compactification of $$X.$$

Let $$T_{X^*}$$ be the topology on $$X^*.$$ Take $$q\not\in X^*$$ and let $$Y=X^*\cup \{q\}.$$ Let the topology on $$Y$$ be $$T_{X^*}\cup \{(U\setminus \{p\})\cup \{q\}: p\in U\in T_{X^*}\}\cup \{Y\}.$$ Then $$Y$$ is compact and id$$_X:X\to Y$$ is a dense homeomorphic embedding.

If $$W$$ is a compact Hausdorff space and if $$f:X\to W$$ is continuous then $$f$$ extends uniquely to a continuous $$f:Y\to W$$ (with $$f(p)=f(q))$$. But $$Y$$ is not Hausdorff.

I can add proofs upon request.