# A "good scale" that is not really a scale

I don't know much about singular cardinal combinatorics, so I apologize in advance if I write something that is wrong or looks funny. First let me recall some basic definitions.

Let $$\lambda$$ be a singular cardinal and let $$(\kappa_i : i < \text{cf}(\lambda))$$ be a sequence of regular cardinals cofinal in $$\lambda$$. A sequence $$(f_\alpha : \alpha < \lambda^+)$$ of functions $$f_\alpha \in \prod_{i < \text{cf}(\lambda)} \kappa_i$$ is called a scale if it is:

(1) increasing modulo $$<^*$$, where $$f <^* g$$ means that $$f(i) < g(i)$$ for all but a bounded set of $$i < \text{cf}(\lambda)$$, and

(2) cofinal in $$\prod_{i < \text{cf}(\lambda)} \kappa_i$$ modulo $$<^*$$.

A scale is said to be good if it satisfies the additional property

(3) for club many $$\alpha < \lambda^+$$, if $$\text{cf}(\alpha) > \text{cf}(\lambda)$$ then $$\alpha$$ is a good point, meaning that there is an unbounded set $$S \subset \alpha$$ and an ordinal $$i < \text{cf}(\lambda)$$ such that the sequence $$(f_\alpha \restriction [i,\text{cf}(\lambda)): \alpha \in S)$$ is pointwise increasing.

The proof of the fact "if $$\lambda$$ is a singular cardinal and there is a strongly compact cardinal between $$\text{cf}(\lambda)$$ and $$\lambda$$, then there is no good scale of length $$\lambda^+$$" seems to use only properties (1) and (3) but not property (2). So my questions are:

(A) Do sequences of length $$\lambda^+$$ satisfying (1) and (3) have a name (e.g. "good pseudo-scale")?

(B) Is their existence equivalent to the existence of good scales?

• If it looks like a scale; and it proves like a scale; it's a duck. Big, red, scaly duck, which looks suspiciously like a dragon. Sep 1, 2014 at 17:27
• (Also, after two votes up, your reputation is so last year! ;-)) Sep 1, 2014 at 17:27

A basic theorem of Shelah: If a $<^*$-increasing sequence of functions has stationary many good points of cofinality $\kappa$ for some $\kappa<\lambda$, then it has an exact upper bound $g$ such that $\mathrm{cf}(g(i))>\kappa$ for but boundedly many $i$.
Taking such an exact upper bound, you can thin to $X\subseteq \mathrm{cf}(\lambda)$ so that $i\mapsto \mathrm{cf}(g(i)): i\in X$ is increasing and unbounded in $\lambda$. Then you can fix cofinal subsets of $g(i)$ of order-type $\mathrm{cf}(g(i))$ for each $i\in X$ and modify the functions to live on $\prod_{i\in X} \mathrm{cf}(g(i))$. You can check that this preserves (3), and the resulting sequence of functions is cofinal in the new product.