## SCH above a compact cardinal

January 20, 2011 at 20:41 | Posted in Uncategorized | 1 Comment
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This is really just an experiment to see if I can actually get latex2wp to work for me, but maybe it has mathematical interest as well. Anyway, it is well-known that Shelah’s pcf theory can be used to give a proof of Solovay’s result that the Singular Cardinals Hypothesis holds above a compact cardinal. What follows is a fairly short proof of a more general result; the argument consists of chaining together some ideas of Shelah taken from his Cardinal Arithmetic.
See “proof of 3.3 in Case gamma” and Remark 3.3B on pages 150-151 of his book.

Theorem 1 Suppose ${\mu}$ is singular of countable cofinality and ${\rm{pp}(\mu)>\mu^+}$. Then there is no uniform countably complete ultrafilter on ${\mu^+}$.

Proof: We actually show more: we prove that there is no weakly ${\mu^+}$-saturated uniform countably complete filter on ${\mu^+}$. Said another way, if ${F}$ is a uniform countably complete filter on ${\mu^+}$ for ${\mu}$ singular of countable cofinality and ${\text{pp}(\mu)>\mu^+}$, then we can find ${\mu^+}$ disjoint ${F}$-positive subsets of ${\mu^+}$. In particular, ${F}$ cannot be an ultrafilter.

To understand the proof, we don’t really need to know what ${\text{pp}(\mu)>\mu^+}$ means. What we really need is one of the combinatorial consequences of this statement, namely that there is a family ${\langle A_\alpha:\alpha<\mu^+\rangle}$ of countable subsets of ${\mu}$ (not ${\mu^+}$!) such that for every ${\beta<\alpha}$, the collection ${\langle A_\beta:\beta<\alpha\rangle}$ is essentially disjoint, in the sense that there is a function ${F_\alpha}$ with domain ${\alpha}$ such that

• ${F_\alpha(\beta)}$ is a finite subset of ${A_\beta}$, and
• the collection ${\langle A_\beta\setminus F_\alpha(\beta):\beta<\alpha\rangle}$ is pairwise disjoint.

Thus, for each ${\alpha<\mu^+}$ the collection ${\langle A_\beta:\beta<\alpha\rangle}$ can be “disjointified” by removing a finite subset from each ${A_\beta}$.

Note that since the ${A_\alpha}$ are subsets of ${\mu}$, no ${\mu^+}$-sized subcollection of ${\langle A_\alpha:\alpha<\mu^+\rangle}$ can be disjointified, as this would give us ${\mu^+}$ disjoint subsets of ${\mu}$. One should think of ${\langle A_\alpha:\alpha<\mu^+\rangle}$ as a failure of “compactness” at ${\mu^+}$ which necessarily arises from the failure of the singular cardinals hypothesis at ${\mu}$.

Let us assume now that ${\langle A_\alpha:\alpha<\mu^+\rangle}$ is as above, and ${F}$ is a uniform countably complete filter on our cardinal ${\mu^+}$; we will show ${\mu^+}$ contains ${\mu^+}$ disjoint ${F}$-positive subsets. Let us also assume that we have fixed some way of enumerating each ${A_\alpha}$, so that expressions such as “the first 5 elements of ${A_\alpha}$” make sense.

Given ${\beta<\mu^+}$ and ${n<\omega}$, define ${X^\beta(n)}$ to be those ${\alpha<\mu^+}$ greater than ${\beta}$ for which ${F_\alpha(\beta)}$ is contained in the first ${n}$ elements of ${A_\beta}$. For each ${\beta}$, the sequence ${\langle X^\beta(n):n<\omega\rangle}$ is increasing with union ${(\beta,\mu^+)}$. Since ${F}$ is uniform and countably complete, it follows that there is an ${n_\beta<\omega}$ such that ${X^\beta(n_\beta)}$ is ${F}$-positive.

For each ${\beta<\mu^+}$, let ${x_\beta}$ be the “${n_\beta+1}$th element” of ${A_\beta}$. Then clearly the set

$\displaystyle Y_\beta:=\{\alpha<\mu^+:\beta<\alpha\text{ and }x_\beta\in A_\beta\setminus F_\alpha(\beta)\} \ \ \ \ \ (1)$

is ${F}$-positive for each ${\beta}$, as includes ${X^\beta(n_\beta)}$.

We know ${x_\beta<\mu}$ for each ${\beta}$, and so there exists an ${x^*<\mu}$ such that

$\displaystyle |\{\beta<\mu^+:x_\beta=x^*\}|=\mu^+. \ \ \ \ \ (2)$

Given ${x^*}$, we define

$\displaystyle Z:=\{\beta<\mu^+:x_\beta=x^*\}. \ \ \ \ \ (3)$

We are now done, as given ${\alpha<\beta}$ in ${Z}$, we see that ${Y_\alpha\cap Y_\beta=\emptyset}$, and therefore the collection ${\{Y_\alpha:\alpha\in A\}}$ is a collection of ${\mu^+}$ disjoint ${F}$-positive subsets of ${\mu^+}$. $\Box$

Corollary 2 If ${\kappa<\mu}$ where ${\kappa}$ is strongly compact and ${\mu}$ is a singular cardinal of countable cofinality, then ${\text{pp}(\mu)=\mu^+}$.

Proof: The co-bounded filter on ${\mu^+}$ is a uniform ${\mu^+}$-complete filter. Since ${\kappa}$ is strongly compact, this filter can be extended to a ${\kappa}$-complete ultrafilter (necessarily uniform) on ${\mu^+}$. The conclusion now follows immediately from the theorem. $\Box$

Corollary 3 If ${\kappa}$ is strongly compact, then the Singular Cardinals Hypothesis holds above ${\kappa}$.

In fact the construction of the sets $A_\alpha$ gives that each one is a cofinal subset of $\mu$ with order type $\omega$, so that “the first five elements of $A_\alpha$” already makes