The Story of Generators – 2October 1, 2012 at 14:22 | Posted in Uncategorized | Leave a comment
Theorem 1 Suppose
- is a progressive sequence of regular cardinals
- is a universal sequence for
- is an exact upper bound for modulo
Then for any ultrafilter on ,
What does this have to do with generators? Just note the following:
Corollary 2 Under the assumptions of the theorem, we have , hence is a generator for .
Proof: If is an ultrafilter on containing , then either meets or it does not. In the first case, the cofinality of is less than by definition of , and in the second case the cofinality is exactly by the conclusion of the theorem. In either case, the cofinality is at most and so is in and we have
For the other inclusion, suppose is in but not in . Let be an ultrafilter on containing but disjoint to . Since the cofinality of is , we have contradicted the conclusion of the theorem.
Digression on exact upper bounds
Before proving Theorem~1, we need to say a few words about exact upper bounds because different authors treat them in slightly different ways. Let us assume that is an exact upper bound for mod just as in the statement of the theorem. It is easy to see
and so our is equal mod to a function satisfying
If we define , then and are equal modulo the ideal , and for any ultrafilter on disjoint to , we have if and only if .
The point of the above is that we can replace by and not change anything, so we may as well assume that our function satisfies
Proof of Theorem
Proof: Suppose first that is an ultrafilter on containing but disjoint to . The sequence is -increasing, so if we can show it is cofinal in we will know that the cofinality of is exactly .
Suppose . Then by setting equal to zero outside of we produce a function that is equal to mod and less than everywhere. Our assumptions on then give us an such
and since , we achieve
For the other direction, suppose by way of contradiction that is an ultrafilter on satisfying with .
Outside of the set , we have and so is -equivalent to a function . Since is universal for , there is an such that and hence
But is an exact upper bound for mod , and so
Since the cofinality of is we know , and therefore
Putting all of this together yields
and this is a contradiction.