## A quick thought

This particular post is tangential to the task at hand. I just want play around with some of the ideas in yesterday’s posts to sharpen them a bit, and maybe prove a fact or two. I’m still not sure if I’m making things too difficult here by bringing in all the pcf stuff; it may be the case that these sorts of facts are well-known and provable by other methods. And once again: I’m writing these entries quickly, just trying to get ideas down, and there’s always the possibility I’m making silly mistakes somewhere. But that’s the point of a blog — I can work out things as I go along, and make corrections where I need to.

In the statement of the following theorem, ${<^*}$ refers to the natural “modulo a bounded set” order, i.e. for ${f}$ and ${g}$ in ${\prod\mathfrak{a}}$, we have ${f<^* g}$ if and only if the set of ${\theta}$ for which ${g(\theta)\leq f(\theta)}$ is bounded in ${\mathfrak{a}}$.

Theorem 1 Let ${\mathfrak{a}}$ be a progressive set of regular cardinals. Then there is a ${\xi<\sup\mathfrak{a}}$ such that

$\displaystyle \max{\rm pcf}(\mathfrak{a}\setminus\xi)={\rm cf}\left(\prod\mathfrak{a}, <^*\right). \ \ \ \ \ (1)$

Recall that

$\displaystyle \max{\rm pcf}(\mathfrak{a}\setminus\xi)={\rm cf}\left(\prod (\mathfrak{a}\setminus\xi), <\right), \ \ \ \ \ (2)$

where “${f” means “${f(\theta) for all ${\theta\in\mathfrak{a}}$”, and so really the theorem gives us a ${\xi<\sup\mathfrak{a}}$ such that

$\displaystyle {\rm cf}\left(\prod (\mathfrak{a}\setminus\xi), <\right)={\rm cf}\left(\prod\mathfrak{a}, <^*\right). \ \ \ \ \ (3)$

Proof: It is clear that

$\displaystyle {\rm cf}\left(\prod\mathfrak{a}, <^*\right)\leq {\rm cf}\left(\prod (\mathfrak{a}\setminus\xi), <\right) \ \ \ \ \ (4)$

for any ${\xi<\sup\mathfrak{a}}$, as a cofinal set in the right hand partial order yields a cofinal set in the left hand one.

Now let us define

$\displaystyle \tau:={\rm cf}\left(\prod\mathfrak{a}, <^*\right)^+ \ \ \ \ \ (5)$

and consider the ideal ${J}$ on ${\mathfrak{a}}$ generated by ${J_{<\tau}[\mathfrak{a}]}$ together with the bounded subsets of ${\mathfrak{a}}$.

We claim that ${\mathfrak{a}\in J}$. Why? Well, assume ${J}$ is a proper ideal on ${\mathfrak{a}}$, and let ${D}$ be an ultrafilter on ${\mathfrak{a}}$ disjoint to ${J}$. Since ${J_{<\tau}[\mathfrak{a}]\subseteq J}$, we know

$\displaystyle {\rm cf}\left(\prod\mathfrak{a}, <^*\right)<\tau\leq{\rm cf}\left(\prod\mathfrak{a}, <_D\right). \ \ \ \ \ (6)$

But ${D}$ is also disjoint to the bounded ideal, so Conclusion 3.2 on page 62 of Cardinal Arithmetic tells us, we have

$\displaystyle {\rm cf}\left(\prod\mathfrak{a}, <_D\right)\leq{\rm cf}\left(\prod\mathfrak{a}, <^*\right) \ \ \ \ \ (7)$

So, it follows that there is a ${\xi<\sup(\mathfrak{a})}$ such that ${\mathfrak{a}\setminus\xi\in J_{<\tau}[\mathfrak{a}]}$. But

$\displaystyle \mathfrak{a}\setminus\xi\in J_{<\tau}[\mathfrak{a}]\Longleftrightarrow\max{\rm pcf}(\mathfrak{a}\setminus\xi)<\tau, \ \ \ \ \ (8)$

and so

$\displaystyle \max{\rm pcf}(\mathfrak{a}\setminus\xi)\leq{\rm cf}\left(\prod\mathfrak{a}, <^*\right) \ \ \ \ \ (9)$

as desired. $\Box$

One obvious corollary of the above is the following result on scales.

Corollary 2 Suppose ${\mu}$ is singular, and ${\langle \mu_i:i<{\rm cf}(\mu)\rangle}$ is an increasing sequence of regular cardinals carrying a scale of length ${\mu^+}$. Then there is an ${i_0<{\rm cf}(\mu)}$ such that ${\langle \mu_i:i_0\leq i<{\rm cf}(\mu)\rangle}$ carries a scale ${\langle f_\alpha:\alpha<\mu^+\rangle}$ with the property that for any ${g\in\prod_{i_0\leq i<{\rm cf}(\mu)}\mu_i}$, there is an ${\alpha<\mu^+}$ such that ${g(i) whenever ${i_0\leq i<{\rm cf}(\mu)}$.

The above is easy to arrange by brute force if ${\mu}$ is of countable cofinality (as the increasing sequence of regular cardinals has ${\max{\rm pcf}}$ equal to ${\mu^+}$) but I don’t see immediately that the brute force argument generalizes to ${\mu}$ of uncountable cofinality, where ${\max{\rm pcf}}$ could be large by virtue of an initial segment of ${\langle \mu_i:i<{\rm cf}(\mu)\rangle}$.