Non-isomorphic free ultrafilters?
Bill Taylor
There are (presumably) vast numbers of different types.
How many? As many as P(R) ? (It can't be more.)
In particular, is there any quick slick way, or even a quick & dirty
way, to show that there can be two free ultrafilters that are NOT
isomorphic, (i.e. with respect to permutations on the naturals).
TIA
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Bill Taylor W.Ta...@math.canterbury.ac.nz
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"Your Eminence, we were told that alcohol is the worst enemy
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fred....@gmail.com
> Consider free ultrafilters on the natural numbers.
>
> There are (presumably) vast numbers of different types.
>
> How many? As many as P(R) ? (It can't be more.)
Yes, as many as P(R).
> In particular, is there any quick slick way, or even a quick & dirty
> way, to show that there can be two free ultrafilters that are NOT
> isomorphic, (i.e. with respect to permutations on the naturals).
Sure. Let c = |R|. There are 2^c distinct ultrafilters on N. If we
partition them into isomorphism classes, there are at most c
ultrafilters in each class (since there are only c permutations of N),
so there are 2^c nonisomorphic ultrafilters. Requiring free
ultrafilters doesn't change anything, since there are only countably
many fixed ultrafilters on N.
Perhaps you wanted a proof of the fact that there are 2^c distinct
ultrafilters on N? (In other words, that the Stone-Cech
compactification of N has cardinality 2^c.) Here goes:
Let {S_n: n in N} be the collection of all subsets of R expressible as
a finite union of intervals with rational endpoints.
For each t in R, define two complementary subsets of N as follows:
X_{t,1} = {n: t in S_n} and X_{t,0} = {n: t not in S_n}.
For each function f:R --> {0,1}, define B_f = {X_{t,f(t)}: t in R}.
Observe that B_f has the finite intersection property, and so it can be
extended to an ultrafilter U_f on N. Clearly, these ultrafilters are
distinct; hence there are as many ultrafilter on N as there are
functions from R to {0,1}, namely 2^c.
fred....@gmail.com
> Consider free ultrafilters on the natural numbers.
>
> There are (presumably) vast numbers of different types.
>
> How many? As many as P(R) ? (It can't be more.)
>
> In particular, is there any quick slick way, or even a quick & dirty
> way, to show that there can be two free ultrafilters that are NOT
> isomorphic, (i.e. with respect to permutations on the naturals).
I already posted the well-known proof that there are 2^c nonisomorphic
ultrafilters on N. Here is an amusing (if correct) example of *two*
nonisomorphic free ultrafilters on N, from some ancient notes of mine.
If U is any free ultrafilter on N, let G(U) be the following game of
length omega between two players, Black and White. At move n, Black
chooses a nonempty finite set B_n subset N, which is required to be
disjoint from his prior choices B_1, ..., B_{n-1}; then White chooses a
point w_n in B_n. Black wins if lim |B_n| = infinity *and* the set
{w_n: n in N} of points chosen by White belongs to U; otherwise White
wins.
My notes say that, in ZFC, one can construct free ultrafilters U_1 and
U_2 such that G(U_1) is a win for Black and G(U_2) is a win for White.
(Of course it follows that U_1 and U_2 can't be isomorphic.)
Unfortunately, those assertions are not accompanied by proofs. The
White-wins case seems easy enough: if U_2 is an ultrafilter on N
containing no sets of natural density 0, then White can win by always
choosing the biggest number in B_n. The construction of U_1 is probably
trickier.
I suppose there should be an ultrafilter U_3 such that G(U_3) is
undetermined, but I can't find anything about that in my notes.