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It is well known that under CH every totally-disconnected compact F-space of weight at most $\omega_1$ can be embedded into the remainder $\omega^*=\beta\omega\setminus\omega$ of the Cech-Stone compactification of $\omega$. My first two questions are as follows:

Question 1: Under which other axioms (or weaker: in which models of set theory) does this fact also hold? What are known consistent counter-examples to this fact?

Question 2: Can this fact be consistently generalised to higher weights, e.g. under GCH?

My last question concerns embedding of some special separable spaces into $\beta\omega$.

Question 3: Let $\kappa$ be an infinite cardinal number and $A$ a closed separable subspace of $\beta\kappa$, where $\kappa$ is endowed with the discrete topology. Can $A$ always be embedded into the space $\beta\omega$? And what in the case when $A$ is countable but not necessarily closed in $\beta\kappa$?

Thank you in advance for the answers or even useful hints!

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    $\begingroup$ Obvious first quesiton: What happens under MA? Can you even get weight at most $\frak c$ in that case? Second obvious question: What happens in the third question when you consider some large cardinals (e.g. ones that admit lots of ultrafilters, like measurable cardinals) instead of just some arbitrary cardinal? $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 12, 2018 at 16:34
  • $\begingroup$ What's going on under MA is a good question, I don't know. And concerning possible properties of $\kappa$, a priori I cannot assume anything. $\endgroup$
    – Damian Sobota
    Commented Dec 12, 2018 at 16:49
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    $\begingroup$ @AsafKaragila: van Douwen and van Mill proved that $\mathsf{MA}+\mathfrak{c} = \aleph_2$ implies there is a compact $F$-space that does not embed in $\beta \omega$. So the answer to your comment is no, $\mathsf{MA}$ does not suffice to give you all spaces of weight at most $\mathfrak{c}$, and in fact it gives you the opposite. jstor.org/stable/1998148?seq=1#metadata_info_tab_contents $\endgroup$
    – Will Brian
    Commented Dec 12, 2018 at 17:03
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    $\begingroup$ @WillBrian: And at the end of the first section they write that the example can be constructed under MA + $\mathfrak{c}=\kappa^+$ for any regular uncountable cardinal. $\endgroup$
    – Damian Sobota
    Commented Dec 12, 2018 at 17:41
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    $\begingroup$ @DamianSobota: Good to know that $\mathfrak{c} = \aleph_2$ isn't strictly necessary. And you're correct that $\omega^*$ is not basically disconnected. But $\beta \omega$ is basically disconnected ("basically disconnected" is a weakening of "extremally disconnected") and $\beta \omega$ and $\omega^*$ embed in each other (which implies that a space embeds in $\omega^*$ if and only if it embeds in $\beta \omega$). $\endgroup$
    – Will Brian
    Commented Dec 12, 2018 at 18:46

1 Answer 1

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Answer to 1: In On closed subspaces of $\omega^*$ (Proc. AMS, 1993) it is shown by Dow, Frankiewicz and Zbierski that in the $\aleph_2$-Cohen model every compact zero-dimensional $F$-space of weight at most $\mathfrak{c}$ is embeddable on $\omega^*$.

Answer to 3: yes. If $A$ and $B$ are countable subsets of $\beta\kappa$ that are separated ($\overline{A}\cap B=\emptyset=A\cap\overline{B}$) then they are contained in disjoint clopen sets; this shows that separable subsets of $\beta\kappa$ are extremally disconnected and hence embeddable in $\beta\omega$. To prove the first claim (which is most likely well known) enumerate $A$ and $B$ as $\{a_n:n\in\omega\}$ and $\{b_n:n\in\omega\}$ respectively. Choose subsets $U_n$ and $V_n$ of $\kappa$ such that $U_n\in a_n$ and $V_n\in b_n$ for all $n$ as well as $U_n\notin b_m$ and $V_n\notin a_m$ for all $m$ and $n$. Next put $X_n=U_n\setminus\bigcup_{i<n}V_i$ and $Y_n=V_n\setminus\bigcup_{i\le n}U_i$. Finally let $X=\bigcup_nX_n$ and $Y=\bigcup_n Y_n$; then $X\cap Y=\emptyset$, and $X\in a_n$ and $Y\in b_n$ for all $n$.

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  • $\begingroup$ Dear KP Hart, thanks! (I noticed your answer only now, sorry.) $\endgroup$
    – Damian Sobota
    Commented Feb 11, 2022 at 11:37

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