\DeclareMathOperator{\pp}{pp}

\def\pcf{\rm{pcf}}

\DeclareMathOperator{\cov}{cov}

\def\cf{\rm{cf}}

\def\REG{\sf {REG}}

\def\restr{\upharpoonright}

\def\bd{\rm{bd}}

\def\subs{\subseteq}

\def\cof{\rm{cof}}

\def\ran{\rm{ran}}

\DeclareMathOperator{\PP}{pp}

\)

Continuing our series of posts, let us recall

- $\cf(\lambda)<\theta=\cf(\theta)<\lambda$,
- $\lambda(1):=\cf_{<\theta}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})$, and
- $\lambda(2)$ 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

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

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

$\lambda(1)+\sup\{\pp_{\Gamma(\theta^+,\theta)}(\lambda^*):\lambda\leq\lambda^*\leq\lambda(1)\wedge\cf(\lambda^*)=\theta\}.$

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

$$\lambda^{+\theta}\leq\lambda(1).$$

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:

$$\cf_{\aleph_0}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})<\cov(\lambda,\lambda,\aleph_1, 2)\Longrightarrow \lambda^{+\omega_1}\leq\cov(\lambda,\lambda,\aleph_1, 2).$$

$$\cf_{\aleph_0}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})<\cov(\lambda,\lambda,\aleph_1, 2)\Longrightarrow \lambda^{+\omega_1}\leq\cov(\lambda,\lambda,\aleph_1, 2).$$

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