Two Point Stone–Čech compactification of an uncountable space

Question

I looked at quite a bit of questions on the site and didn't quite find this but apologies if it's here already.

I was wondering if any one knows if there exists an uncountable space and some Stone–Čech compactification that adds exactly two elements , i.e. $ \lvert \beta X \setminus X \lvert = 2$.

I thought about perhaps starting with $ \Omega_0 = [0, \omega_1) $ and sort of iterating compactifications if that makes sense. First adding $ \omega_1$ to the space which I believe gives us a one point compactification that is the same as the Stone–Čech compactification and then trying to do the same to the space that results from that ( i.e. $\Omega$). The problem is that if we apply the same one point compactification process I believe that since $ \Omega$ is already compact we will end up with a space which has a point isolated. Thus ruling out that the compactification is Stone–Čech .

Perhaps this space is not the one to try and work with. If anyone knows of an example that works or has any ideas that would be appreciated, thanks!

Answer 1

Maybe a disappointingly boring example: take $X = [0, \omega_1) \times \{0,1\}$ with the product topology, i.e. the disjoint sum of two copies of $[0, \omega_1)$. Taking as given the fact that the Stone-Cech compactification (SCc) of $[0, \omega_1)$ is $[0, \omega_1]$, I claim the SCc of $X$ is $Y = [0, \omega_1] \times \{0,1\}$. Clearly $|Y \setminus X| =2$.

To check the details, we note that $Y$ is compact Hausdorff, and show that $Y$ has the appropriate universal property. So suppose $K$ is a compact Hausdorff space and $f : X \to K$ is continuous. We must show $f$ has a unique continuous extension $f' : Y \to K$. The uniqueness is clear because $X$ is dense in $Y$. For the existence, let $f_0, f_1 : [0, \omega_1) \to K$ be defined by $f_i(\alpha) = f(\alpha, i)$; these functions are continuous. So there are unique continuous extensions $f_i' : [0,\omega_1] \to K$. Set $f'(\alpha,i) = f_i'(\alpha)$. Then $f'$ is continuous by the pasting lemma, and for $(\alpha,i) \in X$ we have $f'(\alpha, i) = f_i'(\alpha) = f_i(\alpha) = f(\alpha_i)$, so $f'$ extends $f$.

Similarly, for any finite number $n$, one can take $X = [0, \omega_1) \times n$ and get a space with $|\beta X \setminus X| = n$. (As $[0,\omega_1] \times D$ is not compact for an infinite discrete space $D$, you can't replace $n$ with an infinite cardinal and expect the same thing to work.)

Answer 2

There is an order topology which satisfies this. Simply put two copies of $[0,\omega_1)$ back to back (so the space would look like $(−\omega_1,\omega_1)$. Formally, let $X=2\times \omega_1$ and put the lexicographic ordering on X with the usual ordering reversed in the first factor. So $(i,x)<(j,y)$ in $X$ if $i=0$ and $j=1$, or if $i=j=0$ and $x>y$, or if $i=j=1$ and $x<y$. Endow $X$ with the order topology induced by this ordering. A continuous function on X is eventually constant towards both ends, and so it extends to $[-\omega_1,\omega_1]$, the space obtained by adding first element $\{-\omega_1\}$ and last element $\{\omega_1\}$ to $X$, which is a compactification of $X$. Thus $\beta X=[-\omega_1,\omega_1]$, and $|\beta X\setminus X|=2$.