$\scr{B}$ is the base of a nonprincipal ultrafilter $\scr{U}$ on $\omega$ if 1. $\forall U,V\in\mathscr{B}~\exists T\in\mathscr{B}:~T\subset U\cap V$, 2. $\forall X\in\mathscr{U}~\exists U\in\mathscr{B}: U\subset X$. It is known that there is no countable base for nonprincipal ultrafilter. How we prove this?
1 Answer
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Let $\{U_i\}_{i<\omega}$ be countable base and $A\cap U_i\neq\varnothing$ for all $i<\omega$. Then $A\cap B\neq\varnothing$ for all subsets $B$ from ultrafilter and thus $A\in \scr{U}$. But we can find distinct pairs $\{a_i;b_i\}\in U_i$ for all $i$ because $U_i$ are infinite. For the sets $A=\{a_i\}$ and $B=\{b_i\}$ we have $A\cap U_i\neq\varnothing$, $B\cap U_i\neq\varnothing$. So, $A,B\in\scr{U}$. But $A\cap B=\varnothing$. Contradiction.
