# Embeddability into $\beta\omega$ and $\omega^*$

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 $$\beta\kappa$$?

• 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? – Asaf Karagila Dec 12 '18 at 16:34
• 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. – Damian Sobota Dec 12 '18 at 16:49
• @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 – Will Brian Dec 12 '18 at 17:03
• @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. – Damian Sobota Dec 12 '18 at 17:41
• @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 $\beta \omega$). – Will Brian Dec 12 '18 at 18:46
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 $$X_n=U_n\setminus\bigcup_{i and $$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$$.