## The Story of Generators – 2

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

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

$\displaystyle 0

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

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. $\Box$