Wednesday, June 05, 2013


\def\REG{\sf {REG}}

Continuing our series of posts, let us recall

  1. $\cf(\lambda)<\theta=\cf(\theta)<\lambda$, 
  2. $\lambda(1):=\cf_{<\theta}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})$, and
  3. $\lambda(2)$ is the minimum cardinality of a family $\mathcal{P}\subseteq[\lambda(1)]^{<\lambda}$ such that for any $B\in[\lambda(1)]^\theta$ there is an $A\in\mathcal{P}$ with $A\cap B$ infinite.
I still owe you a proof that

The first of these inequalities requires a bit of machinery, so I'll save that for later and instead give the trivial proof of the second one and then make some comments on the relationship between $\lambda(1)$ and $\lambda(2)$.

Proposition: $\lambda(2)\leq\cov(\lambda(1),\lambda,\theta^+,\theta)$.

Let $\mathcal{P}\subseteq [\lambda(1)]^{<\lambda}$ have the property that any $B\in [\lambda(1)]^{\theta}$ is covered by a union of fewer than $\theta$ sets from $\mathcal{P}$.  Given such a $B$, it follows immediately that there is an $A\in \mathcal{P}$ for which $A\cap B$ is infinite.

Moving on to something a little more interesting, what can we say about the relationship between $\lambda(1)$ and $\lambda(2)$?  

What I would like to prove here is that if $\lambda(1)$ is strictly less than $\lambda(2)$, then there must be a cardinal of cofinality $\theta$ in the interval $(\lambda,\lambda(1)]$.  Said another way, we show
$$\lambda(1)<\lambda^{+\theta}\Longrightarrow \lambda(1)=\lambda(2).$$

There's a cheap proof of this that is easy only by way of quoting  Shelah's cov vs. pp theorem (Theorem 5.4 on page 87 of Cardinal Arithmetic).  This proof works by noting that 
$$\lambda(1)<\lambda(2)\Longrightarrow \lambda(1)<\cov(\lambda(1),\lambda,\theta^+,\theta),$$
and by the cov vs. pp theorem, this latter cardinal is equal to


Since we know $\lambda(1)<\cov(\lambda(1),\lambda,\theta^+,\theta)$, we conclude there must be a singular cardinal of cofinality $\theta$ in the interval $[\lambda,\lambda(1)]$. Since $\cf(\lambda)<\theta$, the only way this can happen is if 

The other proof I have in mind is also easy, and it doesn't require any fancy pcf theory:

For a cardinal $\kappa$, let $P(\kappa)$ be the statement that there is a family $\mathcal{P}\subseteq[\kappa]^<\lambda$ of size $\kappa$ with the property that for any $B\in [\kappa]^\theta$, there is an $A\in\mathcal{P}$ with $A\cap B$ infinite.

Clearly $P(\lambda)$ holds by way of "initial segments" as $\cf(\lambda)<\theta$, and an easy induction shows that $P(\kappa)$ holds if $\lambda\leq\kappa<\lambda^{+\theta}$.  Thus, if $P(\lambda(1))$ fails it must be the case that $\lambda^{+\theta}\leq\lambda(1)$.

Why should we care about the above?  My interest here lies in the question of whether or not the cardinals $\lambda(1)$ and $\cov(\lambda,\lambda,\theta^+,2)$ are actually equal.  What we've shown above is that if they are not, then there's a sizable gap between $\lambda$ and $\cov(\lambda,\lambda,\theta, 2)$.

Cutting down the case where $\lambda$ has size cofinality $\omega$, the above shows:
$$\cf_{\aleph_0}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})<\cov(\lambda,\lambda,\aleph_1, 2)\Longrightarrow \lambda^{+\omega_1}\leq\cov(\lambda,\lambda,\aleph_1, 2).$$

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