# 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  and  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 ), 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?

 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)

 Saharon Shelah: MAD families and SANE players (preprint obtainable from http://arxiv.org/abs/0904.0816)

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.
• So by letting the $A_n$'s (for $n\in\omega$) be subsets of my $b$, I get what I want. Thank you very much. – martin.koeberl May 30 '13 at 10:17
• As a colleague pointed out to me, this does not work completely. If, say $N_1$ and $N_2$ both contain 0, $B_{N_1}$ and $B_{N_2}$ both have $b_0=a_0$ as a subset, hence $c_{N_1}$ might be equal to $c_{N_2}$. I think, one can repair the argument by a better choice of the B_n's. Just let $B_0$ be an infinite proper subset of $A_0$. Then let $B_{n+1}\subsetneq A_{n+1}\setminus(A_0\cup\dots\cup A_n)\cap\bigcap_{i=1}^n(\omega\setminus B_n)$ be infinite (note that this is possible). Then, proceed as above and note that if $a=c_N=c_{N'}$, then $a\subset A_1\cup\dots\cup A_n$ but $a\neq A_i$, contrad. – martin.koeberl May 30 '13 at 13:34