6

Let $X$ be an infinite set, idemmultiple ($|X|^2 = |X|$) if that helps. I am looking for a proof that $X$ has more than $|X|$-many ultrafilters. It has $|X|$-many principal ultrafilters. Without at least some AC I know of no proof that $X$ must have any non-principal ultrafilters. So let us assume BPI, the Prime ideal theorem. If we assume BPI can we prove $|\beta X| > |X|$? ($\beta X$ is of course the set of ultrafilters on $X$). If we have AC then it is standard that $|\beta X| = 2^{2^{|X|}}$. I don't much mind which set theory we work in (the question actually arose in NF, with $X = V$) co's it's the construction (if any) that I am after.

Summary: i am interested in a proof (no theory specified) that BPI implies that an idemmultiple (infinite) set $X$ has more than $|X|$-many ultrafilters

Asaf Karagila
  • 393,674
Thomas Forster
  • 161
  • My hunch is that BPI is so close to AC that the number of ultrafilters is the same under either. – Henno Brandsma Feb 07 '21 at 22:11
  • If I remember correctly, the usual proof of $|\beta X|=2^{2^{|X|}}$ uses AC only in the form of BPI and the statement that $X$ is in bijection with the set of finite subsets of $X$. – Eric Wofsey Feb 07 '21 at 22:12
  • Oh, hrm, the usual proof constructs a surjection from $\beta X$ to $P(P(X))$; getting an injection the other way seems to require more. – Eric Wofsey Feb 07 '21 at 22:19

1 Answers1

5

Recall the usual proof of $|\beta X|=2^{2^{|X|}}$ using AC. To sketch the argument, you replace $X$ with the set $Y$ of pairs $(A,S)$ where $A$ is a finite subset of $X$ and $S$ is a finite set of finite subsets of $X$. Then, you explicitly construct a family of $2^{2^{|X|}}$ pairwise incompatible filters on $Y$, and extend them each to ultrafilters. This gives $2^{2^{|X|}}$ different ultrafilters on $Y$, and hence on $X$ since $|X|=|Y|$.

With a bit of care, you can make this argument still work using only BPI and $|X|^2=|X|$. First, we still have $|X|=|Y|$. To prove this, note that we can totally order $X$ by BPI, and so from $|X|^2=|X|$ we can obtain a family of injections $[X]^n\to X$ for each finite $n$. Also, $|X|^2=|X|$ implies $\aleph_0\leq |X|$ so $|X|\times\aleph_0\leq |X|^2=|X|$. Thus $|X|\leq [X]^{<\omega}\leq|X|\times\aleph_0=|X|$, and so $|Y|=|[X]^{<\omega}\times[[X]^{<\omega}]^{<\omega}|=|X|^2=|X|$.

Now, using BPI, we can extend each of our filters on $Y$ to an ultrafilter. In fact, we can actually extend them simultaneously to get a family of $2^{2^{|X|}}$ different ultrafilters on $Y$, proving that $|\beta X|=|\beta Y|=2^{2^{|X|}}$. More generally, suppose you have a family $(B_i)_{i\in I}$ of nonzero Boolean algebras and you wish to find ultrafilters on all of them simultaneously. Let $B$ be the free product of the $B_i$. Then $B$ is a nonzero Boolean algebra: if we had $1=0$ in $B$, then we would have $1=0$ in the free product of some finitely many $B_i$ (since $B$ is the direct limit of the finite free products), but that is impossible (for instance, by picking ultrafilters on those finitely many $B_i$ to get a homomorphism from their free product to $\{0,1\}$). Thus by BPI, $B$ has an ultrafilter, and it can be pulled back along the canonical homomorphisms $B_i\to B$ to get a family of ultrafilters on each $B_i$. (Thanks to Andreas Blass for providing a variant of this argument in the comments.)

Eric Wofsey
  • 330,363
  • Thanks very much for prompt and helpful response, Eric. Let me chew this over. Thank you. – Thomas Forster Feb 07 '21 at 22:56
  • Forgive my obtuseness, Eric,---and thank you again---but i haven't got my copy of Jech here (I'm out in the jungle). Your set $Y$ is clearly a key feature. What are the multitudinous filters on it that you construct explicitly? – Thomas Forster Feb 07 '21 at 23:18
  • See https://math.stackexchange.com/questions/2999390/how-many-ultrafilters-there-are-in-an-infinite-space for instance. – Eric Wofsey Feb 08 '21 at 00:07
  • 1
    +1. Actually, BPI does let you extend a lot of filters $F_i$ on sets $S_i$ (for $i\in I$) to ultrafilters simultaneously. The idea is to pull the given filters $F_i$ back to filters on $S=\prod_{i\in I}S_i$ and use those to generate a single filter $F$. That is, writing $p_i$ for the projection $S\to S_i$, let $F$ be generated by all the sets $p_i^{-1}(Z)$ for $i\in I$ and $Z\in F_i$. Then if $U$ is an ultrafilter on $S$ extending $F$, its images $p_i(U)$ extend the $F_i$'s. – Andreas Blass Feb 08 '21 at 01:58
  • 1
    @AndreasBlass: Ah, that's very nice. In algebraic terms, if you have a family of nonzero Boolean algebras $B_i$, you can take the free product $B$ of the $B_i$ which will be nonzero, and then an ultrafilter on $B$ gives an ultrafilter on each $B_i$. – Eric Wofsey Feb 08 '21 at 02:09
  • That's nice.... – Thomas Forster Feb 11 '21 at 01:54