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!

  • 1
    $\begingroup$ 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 Dec 12 '18 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 Dec 12 '18 at 16:49
  • 2
    $\begingroup$ @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 Dec 12 '18 at 17:03
  • 1
    $\begingroup$ @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 Dec 12 '18 at 17:41
  • 3
    $\begingroup$ @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 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 $X_n=U_n\setminus\bigcup_{i<n}V_i$ 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 $n$.


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.