Conclusion 3.2A (part 3)

June 17, 2011 at 13:07 | Posted in Uncategorized | Leave a comment

This post is an immediate continuation of the previous one, so all notation etc. carries over from there. I noticed last night that Conclusion 3.2A is really two separate results which have been smashed together. The main part actually looks like it holds without the assumption that {\mu} is not a fixed point, so I’m going to work in a slightly more general context today. If things work out, I’ll formulate “the real theorem” underlying things.

Lemma 1 Let {\mu} be a singular cardinal of cofinality {\kappa}. Then there is a {\tau<\mu} such that

\displaystyle  \sup\{{\rm cf}\left(\prod\mathfrak{b}, <\right):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (1)

Proof: First, note that if we restrict ourselves to {\tau} with {{\rm cf}(\tau)<\tau<\mu}, then any {\mathfrak{b}} relevant to computing the left-hand side of the above equation is necessarily progressive, and hence the cofinality of {\prod\mathfrak{b}} will just be {\max{\rm pcf}(\mathfrak{b})}. Thus, we need only prove that there is a {\tau} in the interval {(\kappa,\mu)} for which

\displaystyle  \sup\{\max{\rm pcf}(\mathfrak{b}):\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}, |\mathfrak{b}|=\kappa\}\leq{\rm pp}(\mu). \ \ \ \ \ (2)

Suppose this is not the case, so for any {\tau<\mu} there is a {\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}} of cardinality {\kappa} with {{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}. The following proposition takes this hypothesis and moves us closer to a contradiction:

Proposition 2 Under the above assumptions, for every {\tau<\mu} there is a {\mathfrak{b}} such that

  • {\mathfrak{b}} is a progressive subset of {(\tau,\mu)\cap{\sf Reg}},
  • {|\mathfrak{b}|\leq\kappa},
  • {{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}, and
  • {\mathfrak{b}} is a bounded subset of {\mu}.

Proof: Without loss of generality, we assume {\kappa<\tau} and so we can find (progressive) {\mathfrak{b}\subseteq (\tau,\mu)\cap{\sf Reg}} of cardinality {\kappa} such that {{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b})}. If {\mathfrak{b}} is bounded below {\mu} then we are done, so assume {\mathfrak{b}} is unbounded.

Let {D} be an ultrafilter on {\mathfrak{b}} such that

\displaystyle  {\rm pp}(\mu)<{\rm cf}\left(\prod\mathfrak{b}/D\right)=\max{\rm pcf}(\mathfrak{b}). \ \ \ \ \ (3)

The above equation, taken together with the fact that {\mathfrak{b}} is unbounded in {\mathfrak{a}} of cardinality {\kappa}, tells us (by way of the definition of {{\rm pp}(\mu)}) that {D} cannot be disjoint to the bounded subsets of {\mathfrak{b}}, i.e,. that there is a {\sigma<\mu} such that

\displaystyle  \mathfrak{b}^*:=\mathfrak{b}\cap\sigma \in D. \ \ \ \ \ (4)

Now this {\mathfrak{b}^*} has all of the required properties. \Box

Armed with the above proposition, we can construct sequences {\langle \mathfrak{b}_i:i<\kappa\rangle} and {\langle \mu_i:i<\kappa\rangle} such that

  • {\langle \mu_i:i<\kappa\rangle} is a strictly increasing sequence of cardinals cofinal in {\mu}
  • {\mu_0=\kappa}
  • {\mathfrak{b}_i\subseteq (\mu_i, \mu_{i+1})\cap{\sf Reg}}
  • {|\mathfrak{b}_i|\leq\kappa} (hence {\mathfrak{b}_i} is progressive)
  • {{\rm pp}(\mu)<\max{\rm pcf}(\mathfrak{b}_i)} for each {i<\kappa}.

Now define

\displaystyle  \mathfrak{b}=\bigcup_{i<\kappa}\mathfrak{b}_i. \ \ \ \ \ (5)

By construction, {\mathfrak{b}} is a progressive set of regular cardinals cofinal in {\mu} of cardinality {{\rm cf}(\mu)}, and so the definition of {{\rm pp}(\mu)} tells us that if {D} is an ultrafilter on {\mathfrak{b}} disjoint to the bounded ideal, then

\displaystyle   {\rm cf}\left(\prod\mathfrak{b}/D\right)\leq{\rm pp}(\mu). \ \ \ \ \ (6)

Let {\sigma={\rm pp}(\mu)^+}, and consider the ideal {J} on {\mathfrak{b}} generated by {J_{<\sigma}[\mathfrak{b}]} and the bounded subsets of {\mathfrak{b}}. It is easy to show that {J} must be a proper ideal on {\mathfrak{b}}: if not, then there would be a {\nu<\mu} such that

\displaystyle  \mathfrak{b}\cap (\nu,\mu)\in J_{<\sigma}[\mathfrak{a}], \ \ \ \ \ (7)

but this is impossible as

\displaystyle  {\rm pp}(\mu)<{\rm pp}(\mu)^+=\sigma\leq\max{\rm pcf}(\mathfrak{b}\cap(\nu,\mu)) \ \ \ \ \ (8)

by our construction of {\mathfrak{b}}. If {D} is an ultrafilter on {\mathfrak{b}} disjoint to {J}, then

\displaystyle  {\rm pp}(\mu)<\sigma\leq{\rm cf}\left(\prod\mathfrak{b}/D\right), \ \ \ \ \ (9)

and we have a contradiction of (6). \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

Create a free website or blog at
Entries and comments feeds.

%d bloggers like this: