TITLE: Embeddability into $\beta\omega$ and $\omega^*$
QUESTION [7 upvotes]: 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!

REPLY [6 votes]: 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 $n$.