Easy Direction

August 29, 2011 at 14:07 | Posted in Uncategorized | Leave a comment

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)<g_\alpha(\eta). \ \ \ \ \ (6)

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

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