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)


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.

  • $\begingroup$ 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. $\endgroup$ – martin.koeberl May 30 '13 at 10:17
  • 2
    $\begingroup$ 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. $\endgroup$ – martin.koeberl May 30 '13 at 13:34

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.