# Uniformization of almost disjoint families

Suppose $$\mathcal{F} \subseteq \mathcal{P} (\omega)$$ is an almost disjoint family and $$\aleph_0 < \vert \mathcal{F} \vert = \kappa < 2^{\aleph_0}$$. Is it consistent that for some such cardinal $$\kappa$$, we can uniformize every two-valued function on the family; that is, if $$\mathcal{F} = \langle A_i \mid i < \kappa \rangle$$ and for each $$i < \kappa$$ the function $$f_i : A_i \to 2$$ is either constant 1 or constant 0, we can find a total function $$f: \omega \to 2$$ that agrees with each function almost everywhere?

Shelah's Proper and Improper Forcing contains a positive result under the assumption that $$\mathcal{F}$$ is a $$\textit{tree}$$. I would like to know if there is something more general, at least for some special $$\kappa$$.

No, this is not consistent: there is (provably in ZFC) an almost disjoint family of size $$\aleph_1$$ and a two-valued function on that family such that the function cannot be uniformized in the way you've described.

To get such a family and function, we'll use a Hausdorff gap, or (more accurately) just an $$(\omega_1,\omega_1)$$-gap. An $$(\omega_1,\omega_1)$$-gap is a double sequence $$\langle A_\alpha :\, \alpha < \omega_1 \rangle$$, $$\langle B_\alpha :\, \alpha < \omega_1 \rangle$$ of subsets of $$\omega$$ such that

(1) the $$A_\alpha$$'s are almost increasing, in the sense that if $$\alpha < \beta$$ then $$A_\alpha \subseteq^* A_\beta$$ (where $$\subseteq^*$$ means that $$A_\alpha \setminus A_\beta$$ is finite).

(2) the $$B_\alpha$$'s are almost decreasing, in the sense that if $$\alpha < \beta$$ then $$B_\alpha \supseteq^* B_\beta$$.

(3) $$A_\alpha \subseteq^* B_\beta$$ for all $$\alpha,\beta < \omega_1$$.

(4) there is no $$C \subseteq \omega$$ such that $$A_\alpha \subseteq^* C \subseteq^* B_\beta$$ for all $$\alpha,\beta < \omega_1$$.

(Actually, there are two different kinds of sequences that are called "Hausdorff gaps" or $$(\omega_1,\omega_1)$$-gaps in various places. Some places use subsets of $$\omega$$ with the $$\subseteq^*$$ relation, as above, and some use functions $$\omega \rightarrow \omega$$ with the $$\leq^*$$ relation instead. The Wikipedia article I linked to takes the latter approach, but the former is more natural for this problem. But let me point out that given a gap in $$(\omega^\omega,\leq^*)$$, we can get one in $$(\mathcal P(\omega),\subseteq^*)$$ by just identifying a function with the set of points in $$\omega \times \omega$$ underneath its graph.)

Given a gap like this, let $$\mathcal A$$ be the family of all sets of the form $$A_{\alpha+1} \setminus A_\alpha$$ or $$B_\alpha \setminus B_{\alpha+1}$$. Let $$f$$ denote the function $$\mathcal A \rightarrow 2$$ that maps every set of the form $$A_{\alpha+1} \setminus A_\alpha$$ to $$0$$ and every set of the form $$B_\alpha \setminus B_{\alpha+1}$$ to $$1$$. Basically, $$(3)$$ implies that this is an AD family, and $$(4)$$ implies that there is no function $$\omega \rightarrow \omega$$ that uniformizes $$f$$. (If we had such a function, the preimages of $$0$$ and $$1$$ would "split the gap" and that can't happen.)

• I was assuming that the sets in an almost disjoint family are infinite. Can you show that $A_{\alpha +1} \setminus A_\alpha$ is infinite? May 28 at 7:22
• @MatteoCasarosa: Yes, Hausdorff's construction gives you that these sets are always infinite. I guess that the definition I wrote down doesn't prescribe this, but you could alter the definition, if you like, by replacing $\subseteq^*$ with $\subset^*$. May 28 at 8:56
• Great! One more question: you have shown how to go from a $\subseteq^\ast$ gap to a $\leq^\ast$ gap. What would the reverse process be? May 28 at 9:23
• @MatteoCasarosa: I'm not sure there is a reverse process. It's not clear to me, anyway, how to produce a $\leq^*$ gap from a $\subseteq^*$ gap. May 29 at 8:54
• Ok but then wouldn't one notion be stronger than the other? May 29 at 20:25