## PCF Hypotheses 2

November 28, 2011 at 12:17 | Posted in Uncategorized | Leave a comment

Almost all of the implications listed at the end of the last post follow immediately from the definitions involved. The one exception to this is the result that the Shelah Weak Hypothesis implies ${|{\rm pcf}\mathfrak{a}|\leq|\mathfrak{a}|}$ for any progressive set of regular cardinals ${\mathfrak{a}}$. It isn’t a difficult result, but it uses “heavy machinery”: the Localization Theorem for pcf.

Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals satisfying ${|\mathfrak{a}|<|{\rm pcf}(\mathfrak{a})|}$. Let ${\mathfrak{b}}$ consist of the first ${|\mathfrak{a}|^+}$ elements of ${{\rm pcf}(\mathfrak{a})}$ greater than ${|\mathfrak{a}|^+}$ (so ${\mathfrak{b}}$ is progressive).

If we define

$\displaystyle \lambda:=\sup\{\kappa_\alpha:\alpha<|\mathfrak{a}|^+\} \ \ \ \ \ (1)$

then clearly ${\lambda}$ is a singular cardinal of cofinality ${|\mathfrak{a}|^+}$. Our goal is to obtain a violation of the Shelah Weak Hypothesis by proving

$\displaystyle |\{\mu<\lambda:\mu\text{ singular and }{\rm pp}(\mu)\geq\lambda\}|\geq |\mathfrak{a}|^+ \ \ \ \ \ (2)$

We mentioned the localization theorem earlier; we will be content just to quote it without proof:

Theorem 1 (Localization Theorem) Suppose ${\mathfrak{a}}$ is a progressive set of regular cardinals and ${\mathfrak{b}\subseteq{\rm pcf}(\mathfrak{a})}$ does not have a weakly inaccessible point of accumulation. Then for any ${\lambda\in{\rm pcf}(\mathfrak{b})}$ there is a set ${\mathfrak{c}\subseteq\mathfrak{b}}$ of cardinality at most ${|\mathfrak{a}|}$ such that ${\lambda\in{\rm pcf}(\mathfrak{c})}$.

The above is Theorem 3.4 on page 337 of The Book. Note that the theorem doesn’t require that ${\mathfrak{b}}$ is progressive; it demands only that ${\mathfrak{b}\cap\kappa}$ is bounded in ${\kappa}$ for every weakly inaccessible cardinal ${\kappa}$ (of course, this is overkill for the application we have in mind).

Given ${\alpha<\lambda}$, note

$\displaystyle \max{\rm pcf}(\mathfrak{b}\cap(\alpha,\lambda))\geq\lambda. \ \ \ \ \ (3)$

(This follows immediately upon consideration of the cofinality of ${\prod\mathfrak{b}/D}$ where ${D}$ is any ultrafilter on ${\mathfrak{b}}$ disjoint to the ideal of bounded subsets of ${\mathfrak{b}}$.)

By the Localization Theorem, it follows that for each ${\alpha<\lambda}$ there is a ${\beta}$ such that ${\alpha<\beta<\lambda}$ and

$\displaystyle \max{\rm pcf}(\mathfrak{b}\cap (\alpha,\beta))\geq\lambda. \ \ \ \ \ (4)$

It follows easily that we can find an increasing and continuous sequence ${\langle\kappa_\alpha:\alpha<|\mathfrak{a}|^+\rangle}$ of cardinals such that

• ${\sup\{\kappa_\alpha:\alpha<|\mathfrak{a}|^+\}=\lambda}$, and
• ${\max{\rm pcf}(\mathfrak{b}\cap (\kappa_\alpha,\kappa_{\alpha+1}))\geq\lambda}$ for each ${\alpha}$.

For ${\alpha<|\mathfrak{a}|^+}$, let us define

$\displaystyle \mathfrak{b}_\alpha=\mathfrak{b}\cap(\kappa_\alpha,\kappa_{\alpha+1}), \ \ \ \ \ (5)$

and let ${D_\alpha}$ be an ultrafilter on ${\mathfrak{b}_\alpha}$ with

$\displaystyle \lambda\leq{\rm tcf}(\prod\mathfrak{b}_\alpha/D_\alpha). \ \ \ \ \ (6)$

Now let ${\mu_\alpha}$ be the least cardinal with

$\displaystyle \mathfrak{b}_\alpha\cap \mu_\alpha\notin D_\alpha, \ \ \ \ \ (7)$

and an easy argument establishes that ${\mu_\alpha}$ is a singular cardinal satisfying ${{\rm pp}(\mu_\alpha)\geq\lambda}$. Since ${\min(\mathfrak{b}_\alpha)\leq\mu_\alpha\leq\sup(\mathfrak{b}_\alpha)}$, it follows that the sequence ${\langle \mu_\alpha:\alpha<|\mathfrak{a}|^+\rangle}$ is strictly increasing, and therefore we have what we need.