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 $\{U\}=\bigcap_{i\in I}K_i?$
Ultrafilters in $\omega$
2 Answers
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