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.

  • $\begingroup$ 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 $X_\alpha$. $\endgroup$ – Not Mike Nov 28 '12 at 23:46
  • $\begingroup$ 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. $\endgroup$ – Mathieu Baillif Nov 29 '12 at 9:41
  • $\begingroup$ 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. $\endgroup$ – KP Hart Sep 13 '18 at 7:37
  • $\begingroup$ @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. $\endgroup$ – Mathieu Baillif Sep 14 '18 at 6:32
  • $\begingroup$ You're right, I misread. $\endgroup$ – KP Hart Sep 14 '18 at 19:44

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.