# Jonsson cardinals and better scales

I want to use this space to sketch the proof of the following known result:

Proposition: If $\mu^+$ is Jónsson for $\mu$ singular, then there is $B\subseteq\mu\cap REG$ cofinal in $\mu$ with $|B|=cf(\mu)$ such that $\prod B$ carries a better scale.

All one needs to do here is show that $pp(\mu)>\mu^+$ and then use that to construct a better scale. So, the Jónsson assumption is moderately silly.

Proof: Let $\lambda=\mu^+$. We begin by showing that $pp(\mu)>\mu^+$. This is pretty standard, so fix a scale $(\vec f, A)$ with $|A|=cf(\mu)$ and we will show that co-boundedly many elements of $A$ must be Jónsson. Otherwise, thin out $A$ so that every element carries a Jónsson cardinal. Let $\theta$ be large enough regular and let $M\prec (H(\theta),\in,<_\theta, \vec f, A)$ be such that $\lambda\in M$, $|M\cap\lambda|=\lambda$, $cf(\mu)+1\subseteq A$ and $\lambda\not\subseteq M$.

It suffices to show that $\sigma\subseteq M$ for each sufficiently large $\sigma<\mu$. So fix $\sigma$, and note that we can find $a\in A$ such that $\sup M\cap a=a$. Otherwise, $Ch_M(a) for all large enough $a$ and so $Ch_M^A\in\prod A/J^{bd}$. But then, there is some $\alpha<\lambda$ with $\alpha\in M$ such that $Ch_M^A<^* f_\alpha$. But this is silly, as $A\subseteq M$ and so there is some $a\in A$ with $Ch_M(a), but $f_\alpha(a)\in M$ as it’s definable from parameters. So now we can find $a>\sigma$ with $a\in M$ and $\sup(M\cap a)=a$, and such that $a$ carries a Jónsson algebra. But then, $a\subseteq M$. As $\sigma<\mu$ was arbitrary, we get a contradiction.

So now we let $A\subseteq \mu$ be such that $|A|=cf(\mu)$, and every $a\in A$ carries a Jónsson cardinal. Then, $\mu^+\notin pcf_{J^{bd}}(A)$, otherwise we could find $B=B_{\mu^+}$ a generator for $J_{<\mu^+}^{J^{bd}}$. In particular, $pcf_{J^{bd}}(B)=\{\mu^+\}$ (since $\min pcf_{J^{bd}}(B)=\mu^+$) and so $tcf(\prod B/J^{bd})=\mu^+$, which is absurd by our earlier argument. Thus, we have that $\min pcf_{J^{bd}}(A)>\mu^+$ and in particular we get that $\prod B/J^{bd}$ is $\mu^{++}$-directed. With this in hand, the rest of the proof is pretty easy.

We begin by fixing a silly square sequence $\bar C=\langle C_\alpha : \alpha<\mu^+\rangle$. We inductively construct a sequence $\langle f_\alpha : \alpha<\mu^+\rangle$ as follows:

Stage $\alpha=0$: Just let $f_0$ be any function in $\prod B$.

Stage $\alpha+1$: Here we let $f_{\alpha+1}>f_\alpha$ be any pointwise larger function in $\prod B$.

Stage $\alpha$ limit: Now we have to do work. For each $c\in C_\alpha$, let $h_c(a)=\sup_{\beta\in c}f_\beta(a)$ if $ot(c)< a$, and $0$ otherwise. Note that these functions are non-zero almost everywhere. Then we let $f_\alpha\in\prod B$ be such that $h_c<^*f_\alpha$ for each $c\in C_\alpha$ (here is where we use the directedness as $|C_\alpha|\leq\mu^+$).

Note that $\vec f=\langle f_\alpha : \alpha<\mu^+$ has an exact upper bound by construction $h$ with the property that $\{a\in A : cf(h(a))<\theta\}$ is bounded for each $\theta<\mu$. First we claim that $\vec f$ is better in $\prod_{a\in A}h(a)$. In other words, we need that:

for every $\gamma$ of the appropriate cofinality, there is a club $C\subseteq \gamma$ such that for every $\alpha_1$ in $C$, there is a $a_{\alpha_1}$ such that $\alpha_0<\alpha_1$ in $C$ implies that $f_{\alpha_0}(a) for every $a>a^*$ in $A$.

This follows directly from the construction, so let $\gamma<\mu^+$ be of cofinality above $cf(\mu)$. Then let $c\in C_\gamma$ be club, and let $\alpha_0<\alpha_1\in acc(C)$. Note then that $c\cap \alpha_1\in C_{\alpha_1}$ so let $ot(c) be such that $h_{c\cap\alpha_1}(a) for all $a>a^*$. In particular, we see that $f_{\alpha_0}(a). So we have betterness.

Then the usual argument shows that (by thinning out $A$ if necessary) $\prod_{a\in A}cf(h(a))$ carries a better scale, which is what we actually want.

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