free ultrafilters on $\mathbb{N}$ 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 $\mathbb{N}$
 A: 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.
