Continuation on pp

February 8, 2011 at 21:02 | Posted in Uncategorized | Leave a comment

The main result of this post is natural, but unsurprising. Oddly enough, though, I can’t find this exact version published anywhere. I mentioned something in this direction in my Handbook of Set Theory argument; I wish I had thought about it a bit more back then, because the proof is certainly easy enough. Anyway, the result is the following:

Theorem 1 The following statements are equivalent for a singular cardinal {\mu}:

  1. {{\rm pp}(\mu)=\mu^+}
  2. If {\mathfrak{a}\subseteq\mu} is a cofinal progressive set of regular cardinals, then {\prod\mathfrak{a}} carries a scale of length {\mu^+} modulo the ideal {J^{\rm bd}[\mathfrak{a}]} of bounded subsets of {\mathfrak{a}}.


We prove first that (2) implies (1), as this implication is almost immediate. Suppose {\mathfrak{a}\subseteq\mu} is a progressive set of regular cardinals cofinal in {\mu} and of cardinality {{\rm cf}(\mu)}. Our assumptions imply that there is a sequence {\langle f_\alpha:\alpha<\mu^+\rangle} of functions from {\prod\mathfrak{a}} such that

  • {\alpha<\beta\rightarrow f_\alpha<f_\beta\mod J^{\rm bd}[\mathfrak{a}]}, and
  • if {g\in\prod\mathfrak{a}}, then there is an {\alpha<\mu^+} such that {g<f_\alpha\mod J^{\rm bd}[\mathfrak{a}]}.

If {D} is any ultrafilter on {\mathfrak{a}} disjoint to {J^{\rm bd}[\mathfrak{a}]}, then the two above statements continue to hold modulo {D}, and hence {\prod\mathfrak{a}/D} has cofinality at most {\mu^+}. But clearly this cofinality must be at least {\mu^+} as {D} extends the co-bounded filter on {\mathfrak{a}}. The statement {{\rm pp}(\mu)=\mu^+} now follows from the definitions involved.

For the other direction, suppose {{\rm pp}(\mu)=\mu^+}, let {\mathfrak{a}} satisfy the assumptions of statement (2), and let {I} denote the ideal {J^{\rm bd}[\mathfrak{a}]}. It is clear that

\displaystyle   \prod\mathfrak{a}/I\text{ is }\mu^+\text{-directed} \ \ \ \ \ (1)

(see Claim 1.3(4) on page 4 of Cardinal Arithmetic), that is, given a collection of functions in {\prod\mathfrak{a}} of cardinality at most {\mu}, there is a single function in {\prod\mathfrak{a}} above them all modulo the ideal {I}.

By Conclusion 3.2 on page 62 of Cardinal Arithmetic, {{\rm cf}\left(\prod\mathfrak{a}/I\right)} is equal to the supremum of all cardinals of the form {{\rm cf}\left(\prod\mathfrak{a}/D\right)}, where {D} is an ultrafilter on {\mathfrak{a}} disjoint to {I}. This latter cardinal is at most {{\rm pp}(\mu)=\mu^+}, and so it follows that

\displaystyle   {\rm cf}\left(\prod\mathfrak{a}/I\right)=\mu^+. \ \ \ \ \ (2)

Given (1) and (2), it is straightforward to construct the desired scale. \Box


Leave a Comment »

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at
Entries and comments feeds.

%d bloggers like this: