[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]

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1 Comment »

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  1. Wow. And people say what I do (maybe did) for a living is hard to understand. 🙂


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