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!

  • 2
    $\begingroup$ Retracting previous comment. You are correct that the one-point and Stone-Cech compactifications of $[0, \omega_1)$ are both equal to $[0, \omega_1]$. This is perhaps easiest to see via the characterization of the S-C compactification as a subspace of $C(X, [0,1])$, together with the fact that real-valued functions on $[0, \omega_1)$ are eventually constant. $\endgroup$ – Nate Eldredge Aug 6 '15 at 5:06
  • 3
    $\begingroup$ So with that in mind, can't we just take two copies of $[0,\omega_1)$? I.e., the space $X = [0, \omega_1) \times \{0,1\}$ with its product topology. It seems clear that the Stone-Cech ought to be $[0, \omega_1] \times \{0,1\}$. It cannot be a one-point compactification, for consider the function $f : X \to \{0,1\}$ defined by $f(\alpha, n) = n$, i.e. send one copy to 0 and the other to 1. Adding a single point at infinity can't give $f$ a continuous extension. Maybe a disappointingly boring example, but I think it works? $\endgroup$ – Nate Eldredge Aug 6 '15 at 5:13
  • 2
    $\begingroup$ @NateEldredge To see that the only (Hausdorff) compactification of $\omega_1$ is the one-point compactification, let $X$ be any compactification, and suppose there are two distinct points $a,b\in X\setminus\omega_1.$. Choose disjoint closed neighborhoods $A,B$ of $a,b$; then $A\cap\omega_1$ and $B\cap\omega_1$ are disjoint closed unbounded subsets of $\omega_1$ which is impossible. $\endgroup$ – bof Aug 6 '15 at 7:38

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.)

  • $\begingroup$ Wow, very nice. Thank you Nate! $\endgroup$ – Jmaff Aug 6 '15 at 9:36

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 $|\beta X\setminus X|=2$.

  • 1
    $\begingroup$ This space is homeomorphic to $[0,\omega_1) \times 2$. $\endgroup$ – user642796 Aug 6 '15 at 7:46
  • $\begingroup$ Yes, it is just another way of looking at the space. $\endgroup$ – Forever Mozart Aug 6 '15 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.