## Easy Direction

My daughter was born last week, so this post has been written by someone for whom sleep is but a fond memory…

Anyway, picking up where we left off, let us define

$\displaystyle \lambda_1:={\rm cov}(\mu,\kappa^+,\theta,\sigma) \ \ \ \ \ (1)$

and

$\displaystyle \lambda_2={\rm cf}^\sigma_{<\theta}\left(\prod({\sf Reg}\cap(\kappa,\mu))\right). \ \ \ \ \ (2)$

We’ll dispose of the easy direction first, and show

$\displaystyle \lambda_2\leq\lambda_1. \ \ \ \ \ (3)$

Thus, let ${\mathcal{P}\subseteq [\mu]^\kappa}$ be a family of cardinality ${\lambda_1}$ with the property that any set in ${[\mu]^{<\theta}}$ is covered by a union of fewer than ${\sigma}$ sets in ${\mathcal{P}}$.

For each ${A\in\mathcal{P}}$, we define a function ${f_A}$ with domain ${{\sf Reg}\cap(\kappa,\mu)}$ by setting

$\displaystyle f_A(\eta)=\sup(A\cap\eta)+1 \ \ \ \ \ (4)$

whenever ${A\cap\eta\neq\emptyset}$; otherwise, set ${f_A(\eta)}$ equal to ${0}$.

Since elements of ${\mathcal{P}}$ have cardinality ${\kappa}$, it is clear that each ${f_A}$ is an element of ${\prod({\sf Reg}\cap(\kappa,\mu))}$. We claim that the collection ${\mathcal{A}=\{f_A:A\in\mathcal{P}\}}$ witnesses ${\lambda_2\leq\lambda_1}$.

This should follow easily from the definitions involved via a proof by contradiction: if not, then there is a set ${B\subseteq {\sf Reg}\cap(\kappa,\mu)}$ of cardinality less than ${\theta}$ and a function ${g\in\prod B}$ such that ${g}$ cannot be majorized by taking the supremum of any collection of fewer than ${\sigma}$ functions in ${\mathcal{A}}$.

By the definition of ${\mathcal{P}}$, we can find ${\sigma_0<\sigma}$ and ${\{A_\alpha:\alpha<\sigma_0\}\subseteq\mathcal{P}}$ such that

$\displaystyle {\rm ran}(g)\subseteq\bigcup_{\alpha<\sigma_0}A_\alpha. \ \ \ \ \ (5)$

For each ${\alpha<\sigma_0}$, let ${g_\alpha}$ denote the function ${f_{A_\alpha}}$. Given ${\eta\in B}$, we know ${g(\eta)}$ lies in ${A_\alpha}$ for some ${\alpha<\sigma_0}$, hence

$\displaystyle g(\eta)\leq\sup(A_\alpha\cap\eta)

Thus, for each ${\eta\in B}$ there is an ${\alpha<\sigma_0}$ such that ${g(\eta), and since each ${g_\alpha}$ is in ${\mathcal{A}}$, we have a contradiction.