almost disjoint ladder system on $\omega_2$

Question

Suppose $\langle s_\alpha : \alpha \in \omega_2 \cap \mathrm{cof}(\omega_1) \rangle$ is a sequence such that each $s_\alpha$ is an increasing cofinal map from $\omega_1$ to $\alpha$. Is it possible that for all $\alpha < \beta$, $\mathrm{ran}(s_\alpha) \cap \mathrm{ran}(s_\beta)$ is finite? Note that a pressing-down plus pigeonhole argument shows that CH must fail.

Answer 1

Following the suggestion, here is the solution, from Baumgartner: Almost-disjoint sets, the dense set problem and the partition calculus, Annals of Math. Logic, 10(1976), p. 424, part 6.

Let $\{A_\xi:\xi<\omega_2\}$ be sets of size $\aleph_1$ with pairwise countable intersection. Set $p\in P$ iff $p$ is a function, $Dom(p)\in [\omega_2]^{<\omega}$, $p(\xi)\in [A_\xi]^{<\omega}$ ($\xi\in Dom(p)$). $p'\leq p$ if $Dom(p')\supseteq Dom(p)$, $p'\supseteq p(\xi)$ and $p(\xi)\cap p(\eta)=p'(\xi)\cap p'(\eta)$ ($\xi\neq\eta\in Dom(p)$). If $G$ is generic, we set $B_\xi=\bigcup\{p(\xi):\xi\in Dom(p),p\in G\}$. Density arguments give $|B_\xi|=\aleph_1$, $|B_\xi\cap B_\eta|<\omega$, essentially the only nontrivial point is to show that $(P,\leq)$ is ccc.

For that, assume that $p_\alpha\in P$ ($\alpha<\omega_1$). By the $\Delta$-system lemma, we can assume that $Dom(p_\alpha)=s\cup s_\alpha$ with $\{s,s_\alpha:\alpha<\omega\}$ disjoint. The common extension of $p_\alpha$, $p_\beta$ will clearly be the coordinatewise union, the only trouble can be if $p_\alpha(\xi)\cap p_\alpha(\eta)\neq p_\beta(\xi)\cap p_\beta(\eta)$ for some pair $\{\xi,\eta\}\in[s]^2$. However, as $|B_\xi\cap B_\eta|\leq\omega$ for $\xi\neq\eta$, there are just countably many possibilities for $p_\alpha(\xi)\cap p_\alpha(\eta)$ for each pair $(\xi,\eta)$, so we can find $\alpha$ and $\beta$ that $p_\alpha(\xi)\cap p_\alpha(\eta)=p_\beta(\xi)\cap p_\beta(\eta)$ for all pairs $\{\xi,\eta\}$ in $s$, and then $p_\alpha$, $p_\beta$ will be compatible.

The nice thing in the argument is that it is iterable, i.e., $|A_\alpha|=\aleph_2$, $|A_\alpha\cap A_\beta|\leq\aleph_1$ for $\alpha<\beta<\omega_3$, then one forcing shrinks each $A_\alpha$ to $A'_\alpha$ so that $|A'_\alpha|=\aleph_2$, $|A'_\alpha\cap A'_\beta|\leq\aleph_0$, then the next forcing thins out $A'_\alpha$ to $A''_\alpha$ such that $A''_\alpha\cap A''_\beta$ is finite, etc.