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}.

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  1. (this is mostly by way of a test to see if I can put maths in comments)

    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
    perfect sense.


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