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)<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)<f_\alpha(a)<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)<f_{\alpha_1}(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)<a^*\in A be such that h_{c\cap\alpha_1}(a)<f_{\alpha_1}(a) for all a>a^*. In particular, we see that f_{\alpha_0}(a)<f_{\alpha_1}(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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