Claim 3.2 page 377

June 15, 2011 at 11:56 | Posted in Uncategorized | Leave a comment

Last time, we left off with the statement of the following theorem:

Theorem 1 Suppose {\mathfrak{a}} is a progressive set of regular cardinals, and

\displaystyle  \aleph_0\leq{\rm cf}(\sigma)=\sigma<\theta\leq|\mathfrak{a}|. \ \ \ \ \ (1)


\displaystyle  {\rm cf}^\sigma_{<\theta}(\prod\mathfrak{a})=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}). \ \ \ \ \ (2)

The above is just Claim 3.2 on page 377 of The Book with the notation changed a little. Today I want to give a proof of this theorem. If you’ve already read through our June 10th post, then today will be simple because it’s essentially the same proof.

Proof: Start by letting {F\subseteq\prod\mathfrak{a}} be {<\sigma}-cofinal in {\prod\mathfrak{b}} for any {\mathfrak{b}\subseteq\mathfrak{a}} of cardinality {<\theta}, and suppose {\kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}. We will show that {\kappa\leq|F|}, thereby establishing the “{\geq}” part of the theorem.

Assume by way of contradiction that {|F|<\kappa}. Let {J} and {\mathfrak{b}} witness that {\kappa} is in {{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}, i.e.,

  • {J} is a {\sigma}-complete ideal on {\mathfrak{a}},
  • {(\prod\mathfrak{a}, <_J)} has true cofinality {\kappa},
  • {|\mathfrak{b}|<\theta}, and
  • {\mathfrak{a}\setminus\mathfrak{b}\in J}.

Since {|F|<\kappa}, there is a function {g\in\prod\mathfrak{a}} such that

\displaystyle  f<_J g\text{ for all}f\in F. \ \ \ \ \ (3)

Since {|\mathfrak{b}|<\theta}, we can find {F_0\subseteq F} of cardinality {<\sigma} such that

\displaystyle   g(\eta)<\sup\{f(\eta):f\in F_0\}\text{ for all }\eta\in\mathfrak{b}. \ \ \ \ \ (4)

Given {f\in F_0}, we define

\displaystyle  A_f:=\{\eta\in\mathfrak{a}: g(\eta)<f(\eta)\}. \ \ \ \ \ (5)

Then (4) tells us

\displaystyle  \mathfrak{b}\subseteq\bigcup\{A_f:f\in F_0\}. \ \ \ \ \ (6)

A couple of things now come to our rescue: since {\mathfrak{a}\setminus\mathfrak{b}\in J} and {J} is {\sigma}-complete, there must be an {f^*\in F_0} such that {A_f} is not in {J}, and this tells us

\displaystyle  \{\eta\in\mathfrak{a}:g(\eta)<f^*(\eta)\}\notin J. \ \ \ \ \ (7)

This is absurd, of course, for {g} was an upper bound of {F} in {(\prod\mathfrak{a}, <_J)}. Thus, {\kappa\leq |F|} and the “{\geq}” part of the theorem has been established.

For the other direction, we once again mirror our June 10th post. Given {\kappa\in{\rm pcf}(\mathfrak{a})}, let {J_{<\kappa}} and {B_{\kappa}} be the associated ideal and generating set. Basic pcf theory tells us

\displaystyle  {\rm tcf}\left(\prod\mathfrak{a}, <_{J_{<\kappa}+(\mathfrak{a}\setminus B_\kappa)}\right)=\kappa, \ \ \ \ \ (8)

and we let {F_\kappa=\{ f^\kappa_\alpha:\alpha<\kappa\}} serve to witness this.

Now define

\displaystyle  F:=\bigcup\{F_\kappa:\kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})\}. \ \ \ \ \ (9)

Clearly {|F|=\sup{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a})}, so we will be done if we can show {F} is {<\sigma}-cofinal in {\prod\mathfrak{b}} for every {\mathfrak{b}\subseteq\mathfrak{a}} of cardinality {<\theta}.

Suppose this is not the case. Then we can find a set {\mathfrak{b}\subseteq\mathfrak{a}} of cardinality {<\theta} and a function {g\in\prod\mathfrak{a}} such that the restriction of {g} to {\mathfrak{b}} cannot be covered by fewer than {\sigma} functions in {F}.

Let us now define a 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\cap\mathfrak{b}\Longrightarrow g(\eta)<\sup\{f(\eta):f\in F_0\}. \ \ \ \ \ (10)

It’s easy to check that {J} is an ideal, and {\mathfrak{a}\setminus\mathfrak{b}\in J} for trivial reasons. Our choice of {g} guarantees that {J} is a {\sigma}-complete proper ideal as well.

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

\displaystyle  I=J+(\mathfrak{a}\setminus B_\kappa). \ \ \ \ \ (11)

The ideal {I} is essentially {J} restricted to {B_\kappa}; it is easy to see that {I} is a proper (by choice of {\kappa}) {\sigma}-complete (as {J} is) ideal on {\mathfrak{a}}. Clearly {\mathfrak{a}\setminus\mathfrak{b}\in I} as well, since this set is in {J}.

The minimality of {\kappa} ensures

\displaystyle  J_{<\kappa}\subseteq J, \ \ \ \ \ (12)

which immediately implies

\displaystyle  J_{<\kappa}+(\mathfrak{a}\setminus B_\kappa)\subseteq I. \ \ \ \ \ (13)

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

\displaystyle  \kappa={\rm tcf}\left(\prod\mathfrak{a}, <_I\right). \ \ \ \ \ (14)

Putting all of the above together, we see

\displaystyle  \kappa\in{\rm pcf}_{\Gamma(\theta,\sigma)}(\mathfrak{a}) \ \ \ \ \ (15)

by way of our ideal {I}.

Our goal now is to reach a contradiction exactly as in the June 10th post by demonstrating that {B_\kappa\in J}, and {B_\kappa\notin J}.

Part of this is trivial, as {\kappa} was selected so that {B_\kappa\notin J}. For the other part, we know that given our choice of {F_\kappa}, there is an {f\in F_\kappa} such that

\displaystyle  g<f\mod J_{<\kappa}+(\mathfrak{a}\setminus B_\kappa), \ \ \ \ \ (16)

which means,

\displaystyle   \{\eta\in B_\kappa: f(\eta)\leq g(\eta)\}\in J_{<\kappa}\subseteq J. \ \ \ \ \ (17)

But since {F_\kappa\subseteq F}, the definition of {J} tells us

\displaystyle   \{\eta\in B_\kappa\cap \mathfrak{b}: g(\eta)<f(\eta)\}\in J. \ \ \ \ \ (18)

Since {\mathfrak{a}\setminus\mathfrak{b}\in J}, we have

\displaystyle   B_\kappa\setminus\mathfrak{b}\subseteq \mathfrak{a}\setminus\mathfrak{b}\in J, \ \ \ \ \ (19)

and now the conjunction of (17), (18), and (19) together with the fact that {J} is an ideal shows us that {B_\kappa\in J}. Contradiction. \Box

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