Spencinar #5: Happy and mad families, part 2

Now we get to Asger's proof.

Definition: If $A$ is an almost disjoint family and $B_0,B_1\subseteq A$, then $B_0$, $B_1$ have uniformly bounded intersection by $k$ if for any $x\in B_0$, $y\in B_1$, $x\cap y\subseteq k$.

Definition: A polarization of $A$ is a countable partition $\{F_n\}_{n\in\omega}$ of $\{(x,y)\in A^2:x\neq y\}$ such that for each $n$, $\mathrm{proj}_0 F_n$ and $\mathrm{proj}_1 F_n$ have uniformly bounded intersection by some $k$.

Here $\mathrm{proj}_0$ and $\mathrm{proj}_1$ are the projections to the left and right coordinates, respectively.

The polarization will help us get to a countable situation where we can diagonalize. We will show the following lemmas:

Lemma 1: $\mathbf{\Sigma}^1_1$ almost disjoint families have polarizations.

Lemma 2: If an almost disjoint family has a polarization, then it is not maximal.

This would prove Mathias's result again.

To prove Theorem 2, we can use Lemma 2 plus the following (which we won't prove here):

Lemma 3: $\mathrm{OCA}_\infty$ + every set of reals has the perfect set property implies that every almost disjoint family has a polarization. ($\mathrm{OCA}_\infty$ is a strengthening of the open coloring axiom)

Lemma 4:  $\mathrm{OCA}_\infty$ holds in Solovay's model.

Proof of Lemma 1: Suppose $T\subseteq [\omega]^{<\omega}\times [\omega]^{<\omega}\times [\omega]^{<\omega}$ is such that $p[T]:=\{(x,y)\in ([\omega]^{\omega})^2: \exists t \, (t,x,y)\textrm{ is a branch of T}\}$ is equal to $\{(x,y)\in A^2:x\neq y\}$ which exists since that set is analytic (this is a basic descriptive set theory representation of an analytic set).

Consider the set of nodes $(\tau, r,s)$ such that every extension $(\tau',r',s')$ has $r'\cap s'=r\cap s$. Given such $(\tau,r,s)$, let $F=p[T_{(\tau,r,s)}]$. Then $\mathrm{proj}_0 F$ and $\mathrm{proj}_1 F$ have uniformly bounded intersection by $\max(r\cap s)$.

We can perform a countable length derivative process, iteratively removing such nodes from $T$. This will terminate in countably many steps (since there are only countably many nodes) and the tree with no such nodes cannot have infinite branches, as this would contradict the almost disjointness of $A$.

Proof of Lemma 2: First we make the following observation: if $A$ is an uncountable almost disjoint family and polarized by $\{F_n\}_{n\in \omega}$, then there is a sequence $B_0,B_1,\ldots$ of uncountable subsets of $A$ such that $B_i$ has uniformly bounded intersection with $\bigcup_{j>i} B_j$ and $B_0=\mathrm{proj}_0 F_n$ for some $n$.

To prove the observation, take $n$ such that $\mathrm{proj}_0 F_n$ and $\mathrm{proj}_1 F_n$ are both uncountable. Take $B_0=\mathrm{proj}_0 F_n$. Now $\mathrm{proj}_1 F_n$ is polarized by $\{F_m\cap (\mathrm{proj}_1 F_n)^2\}_{m\neq n}$. So we can iterate this process for $\omega$ steps, giving the observation.

Suppose $B_i$ are as above. Set $X_{B_i}=\bigcup B_i$. Note that the $X_{B_i}$ form an almost disjoint family, because of the uniform bound on the intersections of sets from $B_i, B_j$ for any given $i,j$.

This process is the first step in an iterative procedure, $A_0=A$, $B_{0,i}=B_i$, $X_{0,i}=X_i$. The iteration will extend transfinitely. At stage $\alpha$, let $A_\alpha$ be $A$ minus the free ideal generated by all of the $X_{\beta,i}$, $\beta<\alpha$. Define $B_{\alpha,i}$ and $X_{\alpha,i}$ in the same way as above. This terminates after countably many steps, since each $B_{\alpha_i}$ is a different $\mathrm{proj}_0 F_n$. Let $\lambda$ be the length of this procedure: the termination condition is that $A_\lambda$ is countable. Now we can use the usual diagonalization to find a set almost disjoint from $A_\lambda$ and each of the $X_{\alpha_i}$. This set will be almost disjoint from every member of $A$, so $A$ cannot be maximal.