Constructing almost disjoint families Let $\mathcal A$ be an almost disjoint family of subsets of $\omega$ and let $\Psi (\mathcal A)$ be the Mrówka space (definition here).
Let $$\mathcal I (\mathcal A)=\{X\subseteq \omega : X\subseteq \bigcup \mathcal B \text { for some finite }\mathcal B\subseteq \mathcal A\}.$$
Say $X\subseteq \omega $ is a partitioner of $\mathcal A$ if $A\cap X=^*\varnothing $ or $A\subseteq ^* X$ for each $A\in\mathcal A$ (here $^*$ means the relation holds for all but finitely many points). Note that if $X$ is a partitioner of $\mathcal A$ then so is $\omega\setminus X$.
A partitioner $X$ is trivial if $X\in \mathcal I (\mathcal A)$ or $\omega\setminus X\in\mathcal I (\mathcal A)$
I think it is a nice (and easy to prove) result that the Stone-Cech remainder $\beta \Psi(\mathcal A)\setminus \Psi(\mathcal A)$ is connected iff $\mathcal A$ has no nontrivial partitioners. My question is: How can we construct almost disjoint families (of subsets of $\omega$) with no nontrivial partitioners?  Are there some basic methods for constructing almost disjoint families that you can share with me?
 A: I’ve seen only one construction. Let $T={}^{<\omega}2$ be the binary tree of height $\omega$. For each $f\in{}^\omega2$ let $B_f=\{f\upharpoonright n:n\in\omega\}$ be the branch determined by $f$, and let $\mathscr{B}=\{B_f:f\in{}^\omega2\}$; $\mathscr{B}$ is an AD family in $\wp(T)$, so it can be extended to a MAD family $\mathscr{A}$. Clearly $|\mathscr{A}\setminus\mathscr{B}|\le 2^\omega$; pick any $F\subseteq{}^\omega2$ such that $|F|=|\mathscr{A}\setminus\mathscr{B}|$, and let $\mathscr{A}\setminus\mathscr{B}=\{A_f:f\in F\}$.
Now let 
$$\mathscr{A}'=\{B_f:f\in{}^\omega2\setminus F\}\cup\{B_f\cup A_f:f\in F\}\;;$$
clearly $\mathscr{A}'$ is a MAD family. Suppose that $X\subseteq{}^{<\omega}2$ is a non-trivial partitioner of $\mathscr{A}'$. Note that $\{A\cap X:A\in\mathscr{A}'\}$ is a MAD family in $\wp(X)$, so it must be uncountable; since $X$ is a partitioner, this means that $F_X=\{f\in{}^\omega2:B_f\subseteq^*X\}$ must be uncountable as well. But
$$F_X=\bigcup_{n\in\omega}\{f\in{}^\omega2:\forall m\ge n(f\upharpoonright m\in X)\}\;,$$
and the sets $\{f\in{}^\omega2:\forall m\ge n(f\upharpoonright m\in X)\}$ are easily seen to be closed in ${}^\omega2$ when the latter is viewed as the product of $\omega$ discrete $2$-point spaces, so $F_X$ is an uncountable $F_\sigma$ in ${}^\omega2$ and must therefore have cardinality $2^\omega$.
Now let $\{X_\xi:\xi<\kappa\}$ be an enumeration of the non-trivial partitioners of $\mathscr{A}'$, where $\kappa\le 2^\omega$. For $\eta<\kappa$ recursively choose distinct $C_\eta,D_\eta\in\mathscr{A}'\setminus\big(\{C_\xi:\xi<\eta\}\cup\{D_\xi:\xi<\eta\}\big)$ so that $C_\eta\subseteq^*X_\eta$ and $D_\eta\cap X_\eta=^*\varnothing$. To see that this is always possible, note that $\eta<2^\omega=|F_X|$, so we can certainly choose $C_\eta$; and ${}^{<\omega}2\setminus X$ is also a partitioner of $\mathscr{A}'$, so for essentially the same reason we can choose $D_\eta$.
Finally, let
$$\mathscr{M}=\{C_\xi\cup D_\xi:\xi<\kappa\}\cup\Big(\mathscr{A}'\setminus\big(\{C_\xi:\xi<\kappa\}\cup\{D_\xi:\xi<\kappa\}\big)\Big)\;;$$
then $\mathscr{M}$ is a MAD family with no non-trivial partitioners: the sets $C_\xi\cup D_\xi$ kill off every potential non-trivial partitioner.
