# Ultrafilters in $\omega$

Let $U$ be a non-principal ultrafilter in $\beta \mathbb{N}$. Can it have a countable character as a point in this topological space? Is there decreasing chain of clopen subsets of $\beta\mathbb{N}$ $(K_i)_{i\in I}$ such that $$\{U\}=\bigcap_{i\in I}K_i?$$

No, there are no non-trivial (i.e. not eventually constant) convergent sequences in $\beta\mathbb{N}$, and a point of countable character that is not isolated (as such a U would be) allows one to define a convergent sequence to it.
For $A\subseteq\Bbb N$ let $\widehat A=\{\mathscr{U}\in\beta\Bbb N:A\in\mathscr{U}\}$.
Let $\mathscr{U}\in\beta\Bbb N\setminus\Bbb N$, and let $\{V_n:n\in\Bbb N\}$ be any countable family of open sets containing $\mathscr{U}$; then $\bigcap_{n\in\Bbb N}V_n\ne\{\mathscr{U}\}$.
To see this, note that for each $n\in\Bbb N$ there must be a $U_n\in\mathscr{U}$ such that $\widehat{U_n}\subseteq V_n$, and we may further assume that $U_{n+1}\subseteq U_n$ for each $n\in\Bbb N$. Now recursively choose distinct $n_k,m_k\in\Bbb N$ for $k\in\Bbb N$ so that $$n_k,m_k\in U_k\setminus\Big(\{n_i:i<k\}\cup\{m_i:i<k\}\Big)\in\mathscr{U}\;;$$ this is possible because $\mathscr{U}$ is non-principal. Let $A=\{n_k:k\in\Bbb N\}$ and $B=\{m_k:k\in\Bbb N\}$, and note that $A\setminus U_n$ and $B\setminus U_n$ are finite for each $n\in\Bbb N$. There are therefore ultrafilters $\mathscr{V}$ and $\mathscr{W}$ on $\Bbb N$ extending the families $\{A\cap U_n:n\in\Bbb N\}$ and $\{B\cap U_n:n\in\Bbb N\}$, respectively. Clearly $U_n\in\mathscr{V}\cap\mathscr{W}$ for each $n\in\Bbb N$, so $\mathscr{V},\mathscr{W}\in\widehat{U_n}$ for each $n\in\Bbb N$, and therefore $\mathscr{V},\mathscr{W}\in\bigcap_{n\in\Bbb N}V_n$. But $A\cap B=\varnothing$, so $\mathscr{V}\ne\mathscr{W}$, and therefore $\bigcap_{n\in\Bbb N}V_n$ is not a singleton.