The Story of Generators – 2

October 1, 2012 at 14:22 | Posted in Uncategorized | Leave a comment

Finding Generators

Theorem 1 Suppose

  • {A} is a progressive sequence of regular cardinals
  • {\lambda\in{\rm pcf}(A)}
  • {\bar{f}=\langle f_\alpha:\alpha<\lambda\rangle} is a universal sequence for {\lambda}
  • {g} is an exact upper bound for {\bar{f}} modulo {J_{<\lambda}[A]}
  • {B=\{\theta\in A:g(\theta)=\theta\}}

Then for any ultrafilter {D} on {A},

\displaystyle  {\rm cf}(\prod A/D)=\lambda \text{ if and only if } B\in D \text{ and }D\cap J_{<\lambda}[A]=\emptyset. \ \ \ \ \ (1)

What does this have to do with generators? Just note the following:

Corollary 2 Under the assumptions of the theorem, we have {J_{\leq\lambda}[A]=J_{<\lambda}[A]+B}, hence {B} is a generator for {\lambda}.

Proof: If {D} is an ultrafilter on {A} containing {B}, then either {D} meets {J_{<\lambda}[A]} or it does not. In the first case, the cofinality of {\prod A/D} is less than {\lambda} by definition of {J_{<\lambda}[A]}, and in the second case the cofinality is exactly {\lambda} by the conclusion of the theorem. In either case, the cofinality is at most {\lambda} and so {B} is in {J_{\leq\lambda}[A]} and we have

\displaystyle  J_{<\lambda}[A]+B\subseteq J_{\leq\lambda}[A]. \ \ \ \ \ (2)

For the other inclusion, suppose {B^*} is in {J_{\leq\lambda}[A]} but not in {J_{<\lambda}[A]+B}. Let {D} be an ultrafilter on {A} containing {B^*} but disjoint to {J_{<\lambda}[A]+B}. Since the cofinality of {\prod A/D} is {\lambda}, we have contradicted the conclusion of the theorem. \Box

Digression on exact upper bounds

Before proving Theorem~1, we need to say a few words about exact upper bounds because different authors treat them in slightly different ways. Let us assume that {g} is an exact upper bound for {\bar{f}} mod {J_{<\lambda}[A]} just as in the statement of the theorem. It is easy to see

\displaystyle  \{\theta\in A: g(\theta)=0\text{ or }\theta<g(\theta)\}\in J_{<\lambda}[A], \ \ \ \ \ (3)

and so our {g} is equal mod {J_{<\lambda}[A]} to a function {g^*} satisfying

\displaystyle  0<g^*(\theta)\leq\theta\text{ for all }\theta\in A. \ \ \ \ \ (4)

If we define {B^*:=\{\theta\in A:g^*(\theta)=\theta\}}, then {B^*} and {B} are equal modulo the ideal {J_{<\lambda}[A]}, and for any ultrafilter {D} on {A} disjoint to {J_{<\lambda}[A]}, we have {B\in D} if and only if {B^*\in D}.

The point of the above is that we can replace {g} by {g^*} and not change anything, so we may as well assume that our function {g} satisfies

\displaystyle  0<g(\theta)\leq \theta\text{ for all }\theta\in A. \ \ \ \ \ (5)

Proof of Theorem

Proof: Suppose first that {D} is an ultrafilter on {A} containing {B} but disjoint to {J_{<\lambda}[A]}. The sequence {\bar{f}} is {<_D}-increasing, so if we can show it is cofinal in {\prod A/D} we will know that the cofinality of {\prod A/D} is exactly {\lambda}.

Suppose {h\in \prod A}. Then by setting {h} equal to zero outside of {B} we produce a function {h^*} that is equal to {h} mod {D} and less than {g} everywhere. Our assumptions on {g} then give us an {\alpha} such

\displaystyle  h^*<_{J_{<\lambda}[A]} f_\alpha, \ \ \ \ \ (6)

and since {D\cap J_{<\lambda}[A]=\emptyset}, we achieve

\displaystyle  h=_D h^*<_D f_\alpha, \ \ \ \ \ (7)

as required.

For the other direction, suppose by way of contradiction that {D} is an ultrafilter on {A} satisfying {{\rm cf}(\prod A/D)=\lambda} with {B\notin D}.

Outside of the set {B}, we have {g(\theta)<\theta} and so {g} is {D}-equivalent to a function {g^*\in\prod A}. Since {\bar{f}} is universal for {\lambda}, there is an {\alpha<\lambda} such that {g^*<_D f_\alpha} and hence

\displaystyle  g<_D f_\alpha \ \ \ \ \ (8)

as well.

But {g} is an exact upper bound for {\bar{f}} mod {J_{<\lambda}[A]}, and so

\displaystyle  f_\alpha\leq_{J_{<\lambda}[A]} g. \ \ \ \ \ (9)

Since the cofinality of {\prod A/ D} is {\lambda} we know {D\cap J_{<\lambda}[A]=\emptyset}, and therefore

\displaystyle  f_\alpha\leq_D g. \ \ \ \ \ (10)

Putting all of this together yields

\displaystyle  f_\alpha\leq_D g <_D f_\alpha, \ \ \ \ \ (11)

and this is a contradiction. \Box

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