Conclusion 3.2A (part 4)

June 17, 2011 at 13:54 | Posted in Uncategorized | Leave a comment

So let us now return to the proof of Conclusion 3.2A. We’ll keep notation the same; what remains to be shown is that if {\mu} is singular of cardinality {\kappa} and {\mu<\aleph_\mu}, then there is a {\tau<\mu} for which

\displaystyle   {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)\leq{\rm pp}(\mu). \ \ \ \ \ (1)

We have assumed {\mu<\aleph_\mu}, so for all sufficiently large {\tau<\mu} we know {(\tau,\mu)\cap{\sf Reg}} is progressive. Given such a {\tau}, we can apply Claim 3.2 (June 15, 2011) to conclude

\displaystyle   {\rm cf}_\kappa\left(\prod(\tau,\mu)\cap{\sf Reg}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg}). \ \ \ \ \ (2)

So we fix can fix {\tau<\mu} large enough so that

  • {\mathfrak{a}:=(\tau,\mu)\cap{\sf Reg}} is progressive,
  • {{\rm cf}_\kappa\left(\prod\mathfrak{a}\right)=\sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}, and
  • {\sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq\mathfrak{a}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu)},

where for the last statement we use the result of the previous posting.

We will get what we need by comparing the last two bullet points:

Suppose {\nu\in{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}(\mathfrak{a})}. This means that we there is an ultrafilter {D} and a set {\mathfrak{b}\subseteq \mathfrak{a}} of cardinality {\leq\kappa} such that

  • {\mathfrak{b}\in D}, and
  • {{\rm cf}(\prod\mathfrak{a}/D)=\nu}.

Clearly we may assume that {\mathfrak{b}} has cardinality exactly {\kappa}, and the second bullet implies

\displaystyle  {\rm cf}(\prod\mathfrak{b}/ D)=\nu \ \ \ \ \ (3)

as well, where we view {D} as an ultrafilter on {\mathfrak{b}}. Chaining things together, we have

\displaystyle  \nu={\rm cf}\left(\prod\mathfrak{a}/D\right)={\rm cf}\left(\prod\mathfrak{b}/D\right)\leq\max{\rm pcf}(\mathfrak{b})\leq{\rm pp}(\mu). \ \ \ \ \ (4)

So for our choice of {\tau}, we have

\displaystyle  \sup{\rm pcf}_{\Gamma(\kappa^+,\aleph_0)}((\tau,\mu)\cap{\sf Reg})\leq{\rm pp}(\mu), \ \ \ \ \ (5)

and so (1) follows from (2).

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: