## [Sh:430.1-4] An easy inequality

January 31, 2011 at 16:38 | Posted in Cardinal Arithmetic, cov vs. pp, [Sh:355], [Sh:430] | 1 Comment
What follows is not really part of the first section of [Sh:430], but it does pin down an important connection between cov and pp. The proposition is a special case of part of Shelah’s “cov vs. pp theorem”, Theorem 5.4 on page 87 of *Cardinal Arithmetic*. Note that there are no special assumptions on here. Getting inequalities in the reverse direction (i.e., showing that covering numbers are less than pp numbers) is generally a more difficult proposition.

Let us recall that is the minimum cardinality of a set such that for any (equivalently ), there is a such that .

Proposition 1If is singular of cofinality , then

*Proof:* Suppose by way of contradiction that

By the definition of , we can find a (not necessarily increasing) sequence of regular cardinals and an ultrafilter on such that

and

Let be a family of cardinality standing as witness to , and let be a -increasing and cofinal sequence of functions in . For each , the range of is a subset of of cardinality at most , and so we can find such that

Since is a regular cardinal greater than , there is a single such that

Thus, (by passing to a subsequence of ) we may as well assume that the range of each is a subset of .

But , and so

by our choice of . Let us now define a function by setting if , and

whenever is in . Our assumptions imply that is in .

If , then for each we have

as .

It follows that is a function in such that for all . This is absurd, given our choice of , and the proof is complete.

[Updated 2-7-11]

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Wow. And people say what I do (maybe did) for a living is hard to understand. 🙂

Comment by Jeff— January 31, 2011 #