A property of completely separable mad families A family of sets $\mathcal{A}\subset[\omega]^\omega$ is called almost disjoint (a.d.) iff $\forall a,b\in\mathcal{A}(a\neq b\rightarrow |a\cap b|<\omega)$ and $\mathcal{A}$ is infinite (as such families turn out to be not that interesting if they are finite).
For an a.d. family $\mathcal{A}$ let $\mathcal{I}(\mathcal{A})$ be the ideal on $\omega$ generated by $\mathcal{A}\cup\{\{n\}:\,n\in\omega\}$ and let $\mathcal{I}^+(\mathcal{A})$ be the corresponding coideal $\mathcal{P}(\omega)\setminus\mathcal{I}(\mathcal{A})$.
An a.d. family is called completely separable (or saturated) if for any $b\in\mathcal{I}^+(\mathcal{A})$  there is $a\in\mathcal{A}$ such that $a\subset b$. It is now easy to see that an infinite completely separable a.d. family must be maximal a.d. (i.e., it is not properly included in another a.d. family) as any $A\subset\omega$ witnessing nonmaximality of $\mathcal{A}$ witnesses that $\mathcal{A}$ is not completely separable.
In [1] and [2] one can read (without proof) that any completely separable a.d. family has the property that for any $b\in\mathcal{I}^+(\mathcal{A})$ there already have to be $2^{\aleph_0}$ many $a\in\mathcal{A}$ s.t. $a\subset b$.
A first try was to take an a.d. family $\mathcal{B}$ on $b$ of size $2^{\aleph_0}$ but since it is not guaranteed that its elements are elements of $\mathcal{I}^+(\mathcal{A})$ one does not get that the elements of $\mathcal{B}$ contain elements of $\mathcal{A}$.
So one way is to find an a.d. family of size $2^{\aleph_0}$ inside $[b]^\omega\cap\mathcal{I}^+(\mathcal{A})$ (which should be possible according to [2]), but I didn't know how this could be done.
Does anybody of you know a proof of this or a reference where one can find a proof?
[1] Dilip Raghavan: A model with no strongly separable almost disjoint families, Israel J. Math., vol. 189 (2012), 39–53. (obtainable from http://www.math.toronto.edu/raghavan/ where it appears as 5 in the papers section)
[2] Saharon Shelah: MAD families and SANE players (preprint obtainable from http://arxiv.org/abs/0904.0816)
 A: Choose distinct sets $A_n\in\mathcal A,A_n\subseteq b$ $(n\in\omega)$. Let $B_n$ be an infinite proper subset of $A_n\setminus(A_0\cup\dots\cup A_{n-1})$. Thus the $B_n$'s are pairwise disjoint, and for $i\in\omega$ we have $A_i\not\subseteq\bigcup\{B_n:n\in\omega\}$. Let $\mathcal N\subset[\omega]^{\omega}$ be an a.d. family with $|\mathcal N|=2^{\aleph_0}$. For each $N\in\mathcal N$, let $B_N=\bigcup\{B_n:n\in N\}$. Then $B_N\in\mathcal I^+(\mathcal A)$, since $B_N$ has infinite intersection with infinitely many elements of $\mathcal A$. For each $N\in\mathcal N$ choose $C_N\in\mathcal A$ so that $C_N\subseteq B_N$. Thus $C_N\ne A_i$ for $i\in \omega$.
I claim that the $C_N$'s are distinct. Assume for a contradiction that $C_N=C_{N'}$ for some $N,N'\in\mathcal N,$ $N\ne N'$. Then $C_N\subseteq B_N\cap B_{N'}=\bigcup\{B_i:i\in N\cap N'\}\subseteq\bigcup\{A_i:i\in N\cap N'\}$. Since $N\cap N'$ is finite, it follows that $C_N\cap A_i$ is infinite for some $i$. Since $C_N,A_i\in\mathcal A$ and $C_N\ne A_i$, this contradicts the fact that $\mathcal A$ is almost disjoint.
Edited to correct an error pointed out in a comment by the original poster.
