## On ADS (Part 2)

February 7, 2012 at 11:11 | Posted in Uncategorized | Leave a comment

So our project is to present a proof of the following result of Shelah:

Theorem 1 Suppose ${\mu}$ is singular and ${{\rm cf}\mu=\aleph_0}$. If ${{\rm cov}(\mu,\mu,\aleph_1, 2)>\mu^+}$, then ${{\rm ADS}_\mu}$ holds.

We first prove a lemma that is intended to make our task a bit easier:

Lemma 2 Suppose ${\mu}$ is a singular cardinal of countable cofinality, and suppose ${A}$ and ${\langle A_\alpha:\alpha<\mu^+\rangle}$ are such that

• ${|A|=\mu}$,
• each ${A_\alpha}$ is a countable subset of ${A}$, and
• for each ${\beta<\mu^+}$, there are finite sets ${B^\beta_\alpha\subseteq A_\alpha}$ such that ${\{A_\alpha\setminus B^\beta_\alpha:\alpha<\mu^+\}}$ is pairwise disjoint.

Then ${{\rm ADS}_\mu}$ holds.

Proof: This is not completely trivial because ${{\rm ADS}_\mu}$ requires sets that are unbounded in ${\mu}$, but it is still an elementary argument. Let ${\pi:A\rightarrow\mu}$ be a bijection, and for each ${\alpha<\mu^+}$ let ${C_\alpha=\pi[A_\alpha]}$. Given ${\beta<\mu^+}$, if we let ${D^\beta_\alpha=\pi[B^\beta_\alpha]}$, then it is certainly the case that

$\displaystyle \{C_\alpha\setminus D^\beta_\alpha:\alpha<\beta\}\text{ is pairwise disjoint.} \ \ \ \ \ (1)$

Now we are almost to ${{\rm ADS}_\mu}$; all we are missing is that the ${C_\alpha}$ need to be unbounded in ${\mu}$. This probably won’t be true in general, but there is an ${\alpha^*<\mu^+}$ so that ${C_\alpha}$ is unbounded in ${\mu}$ whenever ${\alpha^*<\alpha<\mu}$. To see why, suppose this fails. Then we can find ${\theta<\mu}$ so that ${I:=\{\alpha<\mu^+: C_\alpha\subseteq\theta\}}$ is unbounded in ${\mu^+}$.

Now choose ${\beta<\mu^+}$ so that ${|I\cap\beta|=\theta^+}$. Now the collection ${\{C_\alpha\setminus D^\beta_\alpha:\alpha\in I\cap\beta\}}$ gives us a collection of ${\theta^+}$ disjoint subsets of ${\theta}$, and clearly this is absurd.

Thus, if we throw away the first ${\alpha^*}$ sets from our family, the remainder will witness ${{\rm ADS}_\mu}$. $\Box$

I guess the moral of the above is that in the case of singular ${\mu}$, we can safely drop the requirement that each ${A_\alpha}$ is unbounded in ${\mu}$ because this will be true from some point on anyway.

Returning now to the proof of the theorem, our plan is to proceed as follows:

We will define a certain family ${\langle x_\alpha:\alpha<\mu^+\rangle}$ of distinct countable subsets of ${\mu}$. We then set ${\eta_\alpha:\omega\rightarrow x_\alpha}$ to be some bijection, and define

$\displaystyle A_\alpha:=\{\eta_\alpha\upharpoonright\ell:\ell<\omega\}. \ \ \ \ \ (2)$

Since the sets ${x_\alpha}$ are distinct, the collection ${\{A_\alpha:\alpha<\mu^+\}}$ will form an almost disjoint collection of countable subsets of ${A:=^{<\omega}\mu}$, and our aim is to show that this collection satisfies the demands of the preceding lemma. Of course, this proof will depend on choosing the sets ${x_\alpha}$ quite carefully.