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This is exercise $II.10.$ of Kunen's set theory:

Show (in ZFC) that there exist almost disjoint families $\mathcal A,\mathcal B\subseteq P(\omega)$ such that $|\mathcal A|=|\mathcal B-\mathcal A|=\omega_1$ and there is no $d\subseteq \omega$ such that $\forall x\in\mathcal A(|d\cap x|<\omega$) and $\forall x\in \mathcal B-\mathcal A(|x-d|=\omega)$.

Kunen hints:

$\mathcal A=\{a_{\alpha}:\alpha<\omega_1\}$, and $\mathcal B-\mathcal A=\{b_{\alpha}:\alpha<\omega_1\}$. Construct $a_{\alpha}$,$b_{\alpha}$ inductively so that $a_{\alpha}\cap b_{\alpha}=\emptyset$ but $\alpha\neq \beta \rightarrow a_{\alpha}\cap b_{\beta}\neq\emptyset$.

But I have no idea on how to show the hint. All ideas are appreciated.

Thanks

Camilo Arosemena-Serrato
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1 Answers1

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Construct the families recursively. At each stage you’re dealing with a countable family of almost disjoint sets, split into two subfamilies, so the crucial step is this:

Suppose that $\mathscr{A}=\{A_n:n\in\omega\}$, $\mathscr{B}=\{B_n:n\in\omega\}$, and $\mathscr{A}\cup\mathscr{B}\subseteq[\omega]^\omega$ is an almost disjoint family such that $A_n\cap B_n=\varnothing$ for all $n\in\omega$, and $A_m\cap B_n\ne\varnothing$ if $m,n\in\omega$ and $m\ne n$. We want to construct $A,B\in[\omega]^\omega$ so that $A\cap B=\varnothing$, and $A\cap B_n\ne\varnothing\ne A_n\cap B$ for $n\in\omega$.

Suppose that $A=\{a_k:k\in\omega\}$ and $B=\{b_k:k\in\omega\}$, where $a_k<a_{k+1}$ and $b_k<b_{k+1}$ for each $k\in\omega$. We want to choose the $a_k$ and $b_k$ so that

$$a_k\in B_k\setminus\left(\bigcup_{n\le k}A_n\cup\{a_i:i<k\}\cup\{b_i:i<k\}\right)$$

and

$$b_k\in A_k\setminus\left(\bigcup_{n\le k}B_n\cup\{a_i:i\le k\}\cup\{b_i:i<k\}\right)$$

for each $k\in\omega$. This can be done by recursion on $k$, choosing $a_k$ and then $b_k$ at step $k$.

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Brian M. Scott
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