Characterize the sup of pcf sigma-complete

June 10, 2011 at 13:31 | Posted in Uncategorized | Leave a comment

This post should be considered a warmup. To start, I want to recall one of the standard results of pcf theory, namely that if {\mathfrak{a}} is a progressive set of regular cardinal, then

\displaystyle  \max{\rm pcf}(\mathfrak{a})={\rm cf}(\prod\mathfrak{a}, <), \ \ \ \ \ (1)

where the ordering on {\prod\mathfrak{a}} is given by

\displaystyle  f<g\Longleftrightarrow f(\eta)<g(\eta)\text{ for all }\eta\in \mathfrak{a}. \ \ \ \ \ (2)

My aim here is to prove an analogous result characterizing {\sup{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}. For those new to this, a cardinal {\kappa} is in {{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})} if there is a {\sigma}-complete ideal {J} on {\mathfrak{a}} with

\displaystyle   \kappa={\rm tcf}(\prod\mathfrak{a}, <_J), \ \ \ \ \ (3)

which means that there is a {<_J}-increasing {\kappa}-sequence of functions in {\prod\mathfrak{a}} cofinal in {\prod\mathfrak{a}}. In other words, (3) means that {(\prod\mathfrak{a}, <_J)} has cofinality {\kappa}, AND every subset of {\prod\mathfrak{a}} of cardinality less than {\kappa} has an upper bound in {(\prod\mathfrak{a}, <_J)}.

The characterization we give involves the following notion:

Definition 1 Let {\sigma} be a regular cardinal, and suppose {\mathfrak{a}} is a set of regular cardinals. We say that a family {F\subseteq\prod\mathfrak{a}} is {<\sigma}-cofinal in {\prod\mathfrak{a}} if for every {f\in\prod\mathfrak{a}}, there is a family {F_0\subseteq F} of size less than {\sigma} such that {f<\sup F_0}, i.e., such that

\displaystyle  f(\eta)<\sup\{g(\eta):g\in F_0\} \ \ \ \ \ (4)

for all {\eta\in \mathfrak{a}}. The {<\sigma}-cofinality of {\mathfrak{a}} is defined to be the minimum cardinality of a {<\sigma}-cofinal subset of {\mathfrak{a}}. We will denote this cardinal by {{\rm cf}^{\sigma}\prod(\mathfrak{a})}.

So now we come to the promised characterization:

Theorem 2 If {\mathfrak{a}} is a progressive set of regular cardinals, and {\sigma} is a regular cardinal with {\sigma\leq|\mathfrak{a}|}, then

\displaystyle  {\rm cf}^{\sigma}(\prod\mathfrak{a}) = \sup{\rm pcf}_{\sigma{\rm -complete}}(\mathfrak{a}). \ \ \ \ \ (5)

Proof: We prove the {\geq} direction first, so assume {F} is a {<\sigma}-cofinal family in {\prod\mathfrak{a}}, and {\kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}. Let {J} be a {\sigma}-complete ideal on {\mathfrak{a}} such that

\displaystyle   {\rm tcf}(\prod\mathfrak{a}, <_J)=\kappa, \ \ \ \ \ (6)

and assume by way of contradiction that {|F|<\kappa}. Given this, (15) implies that {F} is bounded modulo {J} in {\prod\mathfrak{a}}, so there is a {g\in\prod \mathfrak{a}} with

\displaystyle  f<_J g\text{ for all }f\in F. \ \ \ \ \ (7)

We have assumed, however, that {F} is {<\sigma}-cofinal in {\prod\mathfrak{a}}, so we can fix {\alpha<\sigma} and {F_0=\{f_\beta:\beta<\alpha\}\subseteq F} with

\displaystyle  g<\sup F_0. \ \ \ \ \ (8)

Given {\beta<\alpha}, define

\displaystyle  A_\beta=\{\eta\in\mathfrak{a}: g(\eta)<f_\beta(\eta)\}. \ \ \ \ \ (9)

By our choice of {F_0}, we have

\displaystyle  \mathfrak{a}=\bigcup_{\beta<\alpha}A_\beta. \ \ \ \ \ (10)

Since {J} is {\sigma}-complete and {\alpha<\sigma}, there is a {\beta<\alpha} with {A_\beta\in J^+}. But this is a contradiction as {g} was an upper bound for {F} in {(\prod\mathfrak{a}, <_J)}. Thus, {\kappa\leq |F|} and the inequality “{\geq}” has been established.

For the other direction, we are going to need a little pcf theory. In general, for each {\kappa\in{\rm pcf}(\mathfrak{a})} we have an associated ideal {J_{<\kappa}[\mathfrak{a}]} and generator {B_{\kappa}[\mathfrak{a}]}, and

\displaystyle  {\rm tcf}(\prod\mathfrak{a}, <_{J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa[\mathfrak{a}])})=\kappa. \ \ \ \ \ (11)

This entails the existence of a sequence of functions {\langle f^\kappa_\alpha:\alpha<\kappa\rangle} witnessing the true cofinality of the partial order, and we let {F_\kappa} denote the collection {\{f^\kappa_\alpha:\alpha<\kappa\}}.

Let us define

\displaystyle  F:=\bigcup\{F_\kappa: \kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})\}. \ \ \ \ \ (12)

Clearly {|F|=\sup{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a})}. We claim that {F} is {<\sigma}-cofinal in {\prod\mathfrak{a}}.

Suppose this is not the case. Then there is a function {g\in\prod\mathfrak{a}} that cannot be covered by fewer than {\sigma} functions from {F}. We define an collection {J} of subsets of {\mathfrak{a}} by putting {A} into {J} if and only if there is a family {F_0\subseteq F} of cardinality {<\sigma} such that

\displaystyle  \eta\in A\Longrightarrow g(\eta)<\sup\{f(\eta):f\in F_0\}. \ \ \ \ \ (13)

It is easy to check that {J} is a {\sigma}-complete ideal on {\mathfrak{a}}, and this ideal is proper because of our choice of {g}.

Let {\kappa\in{\rm pcf}(\mathfrak{a})} be minimal with {B_\kappa[\mathfrak{a}]\notin J}, and define

\displaystyle  I= J + (\mathfrak{a}\setminus B_\kappa[\mathfrak{a}]). \ \ \ \ \ (14)

Since {J} is {\sigma}-complete, it follows easily that {I} is as well, and our choice of {\kappa} guarantees that {I} is a proper ideal on {\mathfrak{a}}. Our choice of {\kappa} also guarantees

\displaystyle   J_{<\kappa}[\mathfrak{a}]\subseteq J, \ \ \ \ \ (15)

and hence

\displaystyle  J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa[\mathfrak{a}])\subseteq I. \ \ \ \ \ (16)

Thus, the functions {\langle f^\kappa_\alpha:\alpha<\kappa\rangle} show us

\displaystyle  \kappa={\rm tcf}(\prod\mathfrak{a}, <_I), \ \ \ \ \ (17)

and therefore

\displaystyle  \kappa\in{\rm pcf}_{\sigma{\rm-complete}}(\mathfrak{a}). \ \ \ \ \ (18)

We begin to close in on a contradiction. We know that {J} is a proper ideal on {\mathfrak{a}} and therefore {B_\kappa} (we’ll drop the “{[\mathfrak{a}]}”) from now on) is not in {J}; we will finish the proof by demonstrating that in fact {B_\kappa} MUST be in {J}.

Given the definition of {F_\kappa}, we know there is an {f\in F_\kappa} for which

\displaystyle  g< f\mod J_{<\kappa}[\mathfrak{a}]+(\mathfrak{a}\setminus B_\kappa). \ \ \ \ \ (19)

Said another way,

\displaystyle  \{\eta\in B_\kappa: f(\eta)\leq g(\eta)\}\in J_{<\kappa}[\mathfrak{a}]\subseteq J, \ \ \ \ \ (20)

where the last inclusion is from (15).

But since {F_\kappa\subseteq F}, we know

\displaystyle  \{\eta\in B_{\kappa}: g(\eta)<f(\eta)\}\in J \ \ \ \ \ (21)

as well, by the very definition of {J}.

Thus, {B_\kappa} is the union of two sets in {J}, hence an element of {J}. Contradiction.



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: