## [Sh:430.1-4] An easy inequality

January 31, 2011 at 16:38 | Posted in Cardinal Arithmetic, cov vs. pp, [Sh:355], [Sh:430] | 1 Comment

What follows is not really part of the first section of [Sh:430], but it does pin down an important connection between cov and pp. The proposition is a special case of part of Shelah’s “cov vs. pp theorem”, Theorem 5.4 on page 87 of Cardinal Arithmetic. Note that there are no special assumptions on ${\mu}$ here. Getting inequalities in the reverse direction (i.e., showing that covering numbers are less than pp numbers) is generally a more difficult proposition.

Let us recall that ${{\rm cov}(\mu,\mu,\kappa^+, 2)}$ is the minimum cardinality of a set ${\mathcal{P}\subseteq[\mu]^{<\mu}}$ such that for any ${A\in [\mu]^\kappa}$ (equivalently ${[\mu]^{\leq\kappa}}$), there is a ${B\in\mathcal{P}}$ such that ${A\subseteq B}$.

Proposition 1 If ${\mu}$ is singular of cofinality ${\kappa}$, then

$\displaystyle {\rm pp}(\mu)\leq {\rm cov}(\mu,\mu,\kappa^+,2). \ \ \ \ \ (1)$

Proof: Suppose by way of contradiction that

$\displaystyle \theta:={\rm cov}(\mu,\mu,\kappa^+,2)<{\rm pp}(\mu). \ \ \ \ \ (2)$

By the definition of ${\rm pp}$, we can find a (not necessarily increasing) sequence of regular cardinals ${\langle \mu_i:i<\kappa\rangle}$ and an ultrafilter ${D}$ on ${\kappa}$ such that

$\displaystyle \{i<\kappa:\mu_i\leq\tau\}\notin D\text{ for each }\tau<\mu, \ \ \ \ \ (3)$

and

$\displaystyle \theta<\sigma:={\rm cf}\left(\prod_{i<\kappa}\mu_i, <_D\right). \ \ \ \ \ (4)$

Let ${\mathcal{P}\subseteq [\mu]^{<\mu}}$ be a family of cardinality ${\theta}$ standing as witness to ${{\rm cov}(\mu,\mu,\kappa^+,2)=\theta}$, and let ${\langle f_\alpha:\alpha<\sigma\rangle}$ be a ${<_D}$-increasing and cofinal sequence of functions in ${\prod_{i<\kappa}\mu_i}$. For each ${\alpha<\sigma}$, the range of ${f_\alpha}$ is a subset of ${\mu}$ of cardinality at most ${\kappa}$, and so we can find ${A_\alpha\in \mathcal{P}}$ such that

$\displaystyle {\rm ran}(f_\alpha)\subseteq A_\alpha. \ \ \ \ \ (5)$

Since ${\sigma}$ is a regular cardinal greater than ${\theta}$, there is a single ${A^*\in \mathcal{P}}$ such that

$\displaystyle |\{\alpha<\sigma: A_\alpha=A^*\}|=\sigma. \ \ \ \ \ (6)$

Thus, (by passing to a subsequence of ${\langle f_\alpha:\alpha<\sigma\rangle}$) we may as well assume that the range of each ${f_\alpha}$ is a subset of ${A^*}$.

But ${|A^*|<\mu}$, and so

$\displaystyle B:= \{i<\kappa:|A^*|<\mu_i\}\in D \ \ \ \ \ (7)$

by our choice of ${D}$. Let us now define a function ${g}$ by setting ${g(i)=0}$ if ${i\notin B}$, and

$\displaystyle g(i)=\sup(A^*\cap\mu_i) \ \ \ \ \ (8)$

whenever ${i}$ is in ${B}$. Our assumptions imply that ${g}$ is in ${\prod_{i<\kappa}\mu_i}$.

If ${\alpha<\sigma}$, then for each ${i\in B}$ we have

$\displaystyle f_\alpha(i)\leq g(i) \ \ \ \ \ (9)$

as ${f_\alpha(i)\in A^*\cap\mu_i}$.

It follows that ${g}$ is a function in ${\prod_{i<\kappa}\mu_i}$ such that${f_\alpha\leq_D g}$ for all ${\alpha<\sigma}$. This is absurd, given our choice of ${\langle f_\alpha:\alpha<\sigma\rangle}$, and the proof is complete. $\Box$

[Updated 2-7-11]