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.


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