Back to the task

February 11, 2011 at 21:15 | Posted in Uncategorized | Leave a comment

Returning now to our main project, we extract a lemma from Shelah’s argument and examine it in isolation.

Let {\mathfrak{M}=\langle M_\alpha:\alpha<\mu^+\rangle} and {M^*} be as assumed, and consider the structure

\displaystyle  \left( \prod (\mu\cap{\sf Reg}), <_J\right), \ \ \ \ \ (1)

where {J} denotes the ideal of bounded subsets of {\mu}. It’s pretty easy to see that this structure is {\mu^+}-directed, i.e., that every collection of fewer than {\mu^+} functions in {\prod(\mu\cap{\sf Reg})} has a {<_J}-upper bound.

Thus, we can define a sequence of functions {\langle f_\alpha:\alpha<\mu^+\rangle} in {\prod(\mu\cap{\sf Reg})} by letting {f_\alpha} be the {<_\chi}-least function in {\prod(\mu\cap{\sf Reg})} serving as a {<_J}-upper bound for {M_\alpha\cap\prod(\mu\cap{\sf Reg})}. Using this definition, we achieve

\displaystyle  \langle f_\beta:\beta\leq\alpha\rangle\in M_{\alpha+1} \ \ \ \ \ (2)

for each {\alpha}, and so

\displaystyle  \alpha<\beta<\mu^+\Longrightarrow f_\alpha<_J f_\beta \ \ \ \ \ (3)

as well.

Lemma 1 Let {\mathfrak{a}} be a set of regular cardinals satisfying

  • {\sup(\mathfrak{a})=\mu},
  • {\mathfrak{a}} is progressive, and
  • {\mathfrak{a}\in M^*}.

Then {\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle} is a scale in {\prod\mathfrak{a}} modulo the ideal {J^{\rm bd}[\mathfrak{a}]} of bounded subsets of {\mathfrak{a}}.

Proof: Let {\mathfrak{a}} satisfy the assumptions of the proposition. Since {{\rm pp}(\mu)=\mu^+}, we know that {\prod\mathfrak{a}} carries a scale of length {\mu^+} modulo the ideal {J^{\rm bd}[\mathfrak{a}]}. Since {M^*} is an elementary model of {\mathfrak{A}} and {\mathfrak{a}\in M^*}, it follows that there is such a scale living in {M^*}, say {\bar{f}^{\mathfrak{a}}=\langle f^{\mathfrak{a}}_i:i<\mu^+\rangle}. Note that each {f^{\mathfrak{a}}_i} is in {M^*} as {\mu^+\subseteq M^*}.

Now consider {\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle}. Since {\mathfrak{a}} is unbounded in {\mu}, it follows easily that this sequence is increasing modulo {J^{\rm bd}[\mathfrak{a}]} in {\prod\mathfrak{a}}. Given {i<\mu^+}, there is an {\alpha<\mu^+} with {f^{\mathfrak{a}}_i\in M_\alpha}. But clearly this implies

\displaystyle  f^{\mathfrak{a}}_i< f_\alpha\upharpoonright\mathfrak{a}\mod J^{\rm bd}[\mathfrak{a}]. \ \ \ \ \ (4)

Thus, the sequence {\langle f_\alpha\upharpoonright\mathfrak{a}:\alpha<\mu^+\rangle} is also cofinal in {\prod\mathfrak{a}} modulo {J^{\rm bd}[\mathfrak{a}]}, and we are done. \Box

Notice that the above lemma gives us a way to smuggle information into {M^*}: given {\mathfrak{a}} as in the proposition, it follows that ANY function {g\in\prod\mathfrak{a}}, whether in {M^*} or not, is bounded modulo {J^{\rm bd}[\mathfrak{a}]} by a function of the form {f_\alpha\upharpoonright\mathfrak{a}} from {M^*}. Formulating this as a corollary, we obtain:

Corollary 2 Let {\mathfrak{a}} be a set of regular cardinals as in the preceding lemma. If {g\in\prod\mathfrak{a}}, then there is an {\alpha<\mu^+} such that

\displaystyle  g<f_\alpha\upharpoonright\mathfrak{a}\mod J^{\rm bd}[\mathfrak{a}]. \ \ \ \ \ (5)


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: