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Let $\mathbb{N}$ be the set of natural numbers,
I know the fact that the number of free ultrafilters on $\mathbb{N}$ is uncountable.

I have two questions:

1.How to construct these uncountable free ultrafilters?
2.Does there exist two disjoint free ultrafilters on $\mathbb{N}$

pooja somani
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math112358
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    (2) is trivial : consider a set $A\in U_1$ and a set $B\in U_2$. Then $A\cup B$ must be in both... – Steven Stadnicki Oct 05 '18 at 16:50
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    @StevenStadnicki Or even simpler, each contains $\mathbb{N}$. – Noah Schweber Oct 05 '18 at 16:53
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    If I remember correctly (which I'm not at all sure of), if ZF is consistent then so is ZF + every ultrafilter on $\mathbb{N}$ is principal. If that's the case, that pretty much eliminates any possibility of giving an explicit construction of a free ultrafilter on $\mathbb{N}$. – Daniel Schepler Oct 05 '18 at 17:38
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    @DanielSchepler You are correct. – Noah Schweber Oct 05 '18 at 18:00
  • @DanielSchepler . And the axiom of Choice implies there are $2^{2^k}$ many free ultrafilters on a set of cardinal $k\geq \aleph_0,$ so good luck with a "construction" of the set of all of them – DanielWainfleet Oct 05 '18 at 22:07

1 Answers1

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We can't have disjoint ultrafilters, since every ultrafilter contains $\mathbb{N}$.

Counting the nonprincipal ultrafilters takes a bit more work.

It's easy to show that there are at most $2^{2^{\aleph_0}}$-many free ultrafilters on $\mathbb{N}$, since that's how many sets of sets of natural numbers there are, total.

To show that this is sharp, we construct a family $(X_\eta)_{\eta<2^{\aleph_0}}$ of continuum-many sets of natural numbers which are fully independent: if $A,B\subseteq 2^{\aleph_0}$ are finite and disjoint, then the family $$\{X_\eta: \eta\in A\}\cup\{\overline{X_\delta}: \delta\in B\}$$ has the finite intersection property.

This isn't trivial, but is a bit tedious so I'm omitting the details here.

With such a family in hand, via Zorn's lemma we can show that for any subset $A$ of $2^{\aleph_0}$ there is an ultrafilter $\mathcal{U}$ containing each $X_\eta$ for $\eta\in A$ and not containing any $X_\delta$ for $\delta\not\in A$. This shows that there are at least $2^{2^{\aleph_0}}$-many ultrafilters, and since there are only countably many principal ultrafilters this shows that there are $2^{2^{\aleph_0}}$-many free ultrafilters.

Noah Schweber
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