# Type I subspaces of the Stone Cech compactification of $\omega$

EDIT: I found a construction, see below. I decided not to delete the question in case someone is interested.

A space $X$ is of Type I if $X=\cup_{\alpha<\omega_1} X_\alpha$, where each $X_\alpha$ is open in $X$, $\overline{X_\alpha}\subset X_\beta$ whenever $\alpha<\beta$, and each $\overline{X_\alpha}$ is Lindelöf. Although it is not a part of the usual definition, I assume that Type I spaces are not Lindelöf, so $X\not=X_\alpha$ for any $\alpha$. Many classical spaces of set-theoretic topology are of Type I: $\omega_1$ itself, the long line and the long ray, Aronszajn and Suslin trees with the order topology, etc. However these examples are all first countable, and I am trying to see whether there are Type I subspaces of $\omega^\ast=\beta\omega-\omega$ but have some trouble since I am not at all familiar with this space.

Is there (in ZFC) a Type I subspace of $\omega^\ast$ ?

CONSTRUCTION: There is a countably compact Type I subspace X of $\beta\omega$ such that $X_\alpha$ is Lindelöf and $X_{\alpha+1}$ compact for each $\alpha$.

Given any $p$ in $\omega^\ast$, define $\mathcal{U}=\{U\subset\omega^\ast\\,:\\,U$ is clopen and $p\not\in U\}$. Then $\mathcal{U}$ is a cover of $\omega^\ast -\{p\}$. It is shown in van Mill's article in the Handbook of set theoretic topology (Corollary 1.5.4) that no countable subfamily of $\mathcal{U}$ has a dense union in $\omega^\ast$. I will define $X_\alpha$ by induction, each will be an at most countable union of members of $\mathcal{U}$ and thus none will be dense in $\omega^\ast-\{p\}$.

Set $X_0 = \emptyset$. Given $X_\alpha$, take a finite family $\mathcal{F}\subset\mathcal{U}$ that covers the compact set $\overline{X_\alpha}$ and some $U\in\mathcal{U}$ not included in $\cup\mathcal{F}$. There is such a $U$ because $\cup\mathcal{F}$ is not dense and $\mathcal{U}$ is a cover of $\omega^\ast-\{p\}$. Set $X_{\alpha+1}=U\bigcup\cup\mathcal{F}$. Then $X_{\alpha+1}$ is clopen (and thus compact). For limit $\alpha$ define $X_\alpha=\cup_{\beta<\alpha}X_\beta$, then $X_\alpha$ is Lindelöf ($\sigma$-compact, actually) and not dense in $\omega^\ast-\{p\}$. Take $X=\cup_{\alpha<\omega_1}X_\alpha$, then $X$ is a Type I space which is countably compact since any countably infinite subset lies in some $X_{\alpha+1}$ which is compact, so there is an accumulation point.

• K. Kunen proved that there are always weak P-points in $\omega^{\ast}$, (Note: a weak P-point is a point $p$, which is not in the closure of any-countable subset of $\omega^{\ast}\backslash \{p\}$ . This should be enough to make a modified version of your argument work, where you use nested countable subsets of $\omega^{\ast}$ whose closures avoid the weak P-point, to build the sequence $X_\alpha$. – Not Mike Nov 28 '12 at 23:46
• Thanks for your suggestion. I thought of using weak P-points but I did not see how to ensure that $\overline{X_{\alpha}}\subset X_{\alpha+1}$ and still have separable $X_\alpha$'s. Can it be done ? However I think I found a construction, see my EDIT above. – Mathieu Baillif Nov 29 '12 at 9:41
• A Suslin line is not Type I: the family $\{X_{\alpha+1}\setminus\overline{X}_\alpha:\alpha < \omega_1\}$ consists of nonempty pairwise disjoint open sets and is uncountable. – KP Hart Sep 13 '18 at 7:37
• @KP Hart: Yes, a Suslin line is not of Type I, but a Suslin tree with the order topology is. I do not claim it for Suslin lines. – Mathieu Baillif Sep 14 '18 at 6:32
• You're right, I misread. – KP Hart Sep 14 '18 at 19:44