Is there a diagonal argument to show that if $x$ is infinite then ${\cal P}(x)$ (the power set of $x$) is smaller than $\beta x$ (the set of ultrafilters on $x$)?
(Added later. I tried commenting but it wouldn't let me!)
My reason for interest in this was roughly as follows. With AC one can prove that an infinite set $X$ has $2^{2^{|X|}}$ ultrafilters. If choice fails very badly then there might only be $|X|$-many. I was hoping that there might be a diagonal construction lurking in the background which can feed off the choice principles one supplies. As my correspondents know, i spend a lot of time thinking about Quine's NF, and one question there is: how many ultrafilters are there on the universe? If every ultrafilter is principal then very few, yes. But if there are nonprincipal ufs at all, is there some elementary argument available to show that there must be as many of them as there are sets? If there is such a diagonal construction i'd like to know. It could be useful paedogogically, too.
Best Answer
I think there can be no such argument, for the following reason.
It is consistent with ZF that the statement isn't true, since $x$ might be an amorphous set, a set for which every subset is either finite or co-finite. In this case, every ultrafilter either contains a finite set, in which case it must be the principal ultrafilter concentrating on a point of $x$, or else it is exactly the cofinite filter, which is an ultrafilter when $x$ is amorphous. Thus, the number of ultrafilters on an amorphous set $x$ is precisely $x+1$, which is strictly smaller than $2^x$.
So any proof of the statement would involve a choice principle, and this would seem to indicate that it couldn't be a purely diagonal argument.