## The Story of Generators – 3 (Existence)

The Exact Upper Bound Argument

Let us assume that ${A}$ is as usual, with ${\lambda\in{\rm pcf}(A)}$. If we want to produce a generator for ${\lambda}$, we saw last time that what we need is a universal sequence for ${\lambda}$ that in addition possesses an exact upper bound modulo ${J_{<\lambda}[A]}$.

I don’t want to make an excursion into the theory of exact upper bounds, as there have been many high quality write-ups of this material already: in addition to Section 2.1 of the Abraham/Magidor article, there are also some materials due to Kojman that do an excellent job of exposing this material. The point is that one can use results of Shelah to modify the universal sequence so that it ends up with an exact upper bound, and then we get generators by the previous post.

Instead of re-hashing such arguments, I offer the following direct route from universal sequences to generators. The proof takes advantage of the ideal ${I[\lambda]}$, and we will see similar arguments shortly when we talk about transitivity of generators. The argument is based on the exposition of generators given in the Burke-Magidor pcf paper. It turns out that framing things in terms of ${I[\lambda]}$ simplifies things considerably.

From universal sequences to generators

Theorem 1 Suppose ${A}$ is a progressive set of regular cardinals and ${\lambda\in{\rm pcf}(A)}$. Then there is a generator for ${\lambda}$.

Proof: We can assume that ${\lambda>|A|^{++}}$. (If either ${|A|^+}$ or ${|A|^{++}}$ are in ${A}$, then they have generators consisting of singletons.)

Let ${\bar{f}=\langle f_\alpha:\alpha<\lambda\rangle}$ be a universal sequence for ${\lambda}$, that is, a sequence such that

• ${\alpha<\beta\Longrightarrow f_\alpha<_{J_{<\lambda}[A]} f_\beta}$, and
• ${\langle f_\alpha:\alpha<\lambda\rangle}$ is cofinal in ${\prod A/D}$ whenever ${D}$ is an ultrafilter on ${A}$ satisfying ${{\rm cf}(\prod A/D)=\lambda}$.

Such sequences exist by Theorem 4.2 of the Abraham/Magidor article, or see Lemma 2.1 on page 327 of Cardinal Arithmetic.

Let ${\kappa=|A|^+}$. Since we have assumed ${\kappa^+<\lambda}$, we know there is a stationary set ${S\subseteq S^\lambda_\kappa}$ lying in the ideal ${I[\lambda]}$. (See the first section of [Sh:420], or Theorem 3.18 in my own Handbook article.)

This means that there is a family ${\bar{C}=\langle C_\alpha:\alpha<\lambda\rangle}$ and a club ${E\subseteq\lambda}$ such that

• ${C_\alpha}$ is a closed (possibly bounded) subset of ${\alpha}$
• if ${\beta\in{\rm nacc}(C_\alpha)}$ then ${C_\beta=C_\alpha\cap\beta}$, and
• if ${\delta\in E\cap S}$ then ${C_\delta}$ is club in ${\delta}$ of order-type ${\kappa}$.

For each ${\alpha<\lambda}$ with ${C_\alpha\neq\emptyset}$, we define

$\displaystyle f^*_\alpha=\sup\{f_\beta:\beta\in C_\alpha\}. \ \ \ \ \ (1)$

If ${C_\alpha=\emptyset}$, we set ${f^*_\alpha=f_\alpha}$.

Since ${\prod A/J_{\leq\lambda}[A]}$ is ${\lambda^+}$-directed, we can find a single function ${h}$ bounding the collection ${\{f^*_\alpha:\alpha<\lambda\}}$ modulo ${J_{\leq\lambda}[A]}$.

Now we let ${N}$ be an elementary submodel of ${H(\chi)}$ for some sufficiently large regular cardinal ${\chi}$ such that

• ${A}$, ${\bar{f}}$, ${\bar{C}}$, ${E}$, ${S}$, ${h}$, and ${\lambda}$ are all in ${N}$
• ${|N|<\lambda}$
• ${N\cap\lambda}$ is an ordinal ${\delta\in S}$

This is possible because ${S}$ is stationary in ${\lambda}$.

We define

$\displaystyle B:=\{\theta\in A: h(\theta)

We prove that ${B}$ is a generator for ${\lambda}$. Part of this is simple, as by choice of ${h}$ we know ${B\in J_{\leq\lambda}[A]}$. To finish, we must establish the following:

Proposition 2 If ${D}$ is an ultrafilter on ${A}$ with ${{\rm cf}(\prod A/D)=\lambda}$, then ${B\in D}$.

The proof is not difficult, but the following lemma is critical.

Lemma 3 ${B\in N}$.

Proof:

This is where the ${I[\lambda]}$ assumption pays dividends. Note that ${\delta\in E}$ as ${E\in N}$, and therefore we know that ${C_\delta}$ is club in ${\delta}$ with order-type ${\kappa}$.

Given ${\alpha<\beta}$ in ${{\rm nacc}(C_\delta)}$, we know that ${C_\alpha= C_\delta\cap\alpha}$ is an initial segment of ${C_\beta= C_\delta\cap\beta}$, and both are initial segments of ${C_\delta}$. Thus

$\displaystyle f^*_\alpha\leq f^*_\beta\leq f^*_\delta. \ \ \ \ \ (3)$

Given ${\theta\in A}$ and ${\xi, we know there is a ${\gamma\in C_\delta}$ with ${\xi. Since ${C_\delta}$ is cofinal in ${\delta}$, we know ${\min(C_\delta\setminus(\gamma+1))}$ is an ordinal ${\alpha}$ in ${{\rm nacc}(C_\delta)}$ and ${\gamma\in C_\alpha= C_\delta\cap\alpha}$. Thus ${\xi, and we see

$\displaystyle f^*_\delta=\sup\{f^*_\alpha:\alpha\in{\rm nacc}(C_\delta)\}. \ \ \ \ \ (4)$

In particular, if ${h(\theta), then there is an ${\alpha(\theta)<\delta}$ such that

$\displaystyle \alpha\in {\rm nacc}(C_\delta)\setminus\alpha(\theta)\Longrightarrow h(\theta)

Since ${C_\delta}$ is club in ${\delta}$ with order-type ${\kappa>|A|}$, we can find a single ${\alpha\in{\rm nacc}(C_delta)}$ such that for all ${\theta\in A}$,

$\displaystyle h(\theta)

In particular,

$\displaystyle B=\{\theta\in A: h(\theta)

The set on the right is definable from parameters available in ${N}$ (both ${h}$ and ${f^*_\alpha}$ are there), and therefore ${B\in N}$. $\Box$

Let us return now to the proof of Proposition 2.

Proof: By way of contradiction, suppose ${D}$ is an ultrafilter on ${A}$ forming a counterexample. Since ${B}$ is in ${N}$, we can assume that ${D}$ is in ${N}$ as well (this is the key point).

This sequence ${\bar{f}}$ is universal for ${\lambda}$ and therefore the collection ${\{f_\alpha:\alpha<\lambda\}}$ is cofinal in ${\prod A/D}$. Since ${D\cap J_{<\lambda}[A]=\emptyset}$, we know furthermore that the sequence ${\langle f_\alpha:\alpha<\lambda\rangle}$ is ${<_D}$ increasing.

Thus, there is an ${\alpha^*<\lambda}$ such that ${h<_D f_\alpha}$ whenever ${\alpha^*\leq\alpha<\lambda}$. Again, we may assume that ${\alpha^*}$ is in ${N}$ (and hence less than ${\delta=N\cap\lambda}$).

Choose ${\beta\in C_\delta}$ greater than ${\alpha^*}$ and let ${\alpha=\min(C_\delta\setminus(\beta+1)}$. Then ${\alpha\in{\rm nacc}(C_\delta)}$ and ${C_\alpha= C_\delta\cap\alpha}$. In particular, ${\alpha\in N}$ and ${\beta\in C_\alpha}$

So we have

$\displaystyle h<_D f^*_\alpha\leq f^*_\delta. \ \ \ \ \ (8)$

One the other hand, since ${B\notin D}$ we know

$\displaystyle f^*_\delta\leq_D h. \ \ \ \ \ (9)$

But now we have obtained a contradiction. $\Box$

Proposition 2 taken together with the fact that ${B\in J_{\leq\lambda}[A]}$ easily establishes that ${B}$ is a generator for ${\lambda}$, so we are done. $\Box$