## A quick thought

April 26, 2011 at 11:13 | Posted in Uncategorized | Leave a commentThis particular post is tangential to the task at hand. I just want play around with some of the ideas in yesterday’s posts to sharpen them a bit, and maybe prove a fact or two. I’m still not sure if I’m making things too difficult here by bringing in all the pcf stuff; it may be the case that these sorts of facts are well-known and provable by other methods. And once again: I’m writing these entries quickly, just trying to get ideas down, and there’s always the possibility I’m making silly mistakes somewhere. But that’s the point of a blog — I can work out things as I go along, and make corrections where I need to.

In the statement of the following theorem, refers to the natural “modulo a bounded set” order, i.e. for and in , we have if and only if the set of for which is bounded in .

Theorem 1Let be a progressive set of regular cardinals. Then there is a such that

Recall that

where “” means “ for all ”, and so really the theorem gives us a such that

*Proof:* It is clear that

for any , as a cofinal set in the right hand partial order yields a cofinal set in the left hand one.

Now let us define

and consider the ideal on generated by together with the bounded subsets of .

We claim that . Why? Well, assume is a proper ideal on , and let be an ultrafilter on disjoint to . Since , we know

But is also disjoint to the bounded ideal, so Conclusion 3.2 on page 62 of *Cardinal Arithmetic* tells us, we have

and we have a contradiction.

So, it follows that there is a such that . But

and so

as desired.

One obvious corollary of the above is the following result on scales.

Corollary 2Suppose is singular, and is an increasing sequence of regular cardinals carrying a scale of length . Then there is an such that carries a scale with the property that for any , there is an such that whenever .

The above is easy to arrange by brute force if is of countable cofinality (as the increasing sequence of regular cardinals has equal to ) but I don’t see immediately that the brute force argument generalizes to of uncountable cofinality, where could be large by virtue of an initial segment of .

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