0
$\begingroup$

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 $$\{U\}=\bigcap_{i\in I}K_i?$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.