given a set closed under finite complementation and union; disprove closeness under countable union and intersection The collection $\mathscr{A}$ of subsets of $\Omega=\\{0,1\\}^{\infty}$ is given by $$\mathscr{A}=\\{S\times\Omega:S\subseteq\\{0,1\\}^{l},l\ge1\\}$$ Then, $\mathscr{A}$ is closed under complementation, finite union and hence, finite intersection. > If $A_1,A_2...\in\mathscr{A}$, is > > $(i)$ ${\underset{n=1}{\stackrel{\infty}{\bigcup}}}A_n\in\mathscr{A}$ ? > $(ii)$ ${\underset{n=1}{\stackrel{\infty}{\bigcap}}}A_n\in\mathscr{A}$ ? Somehow, intuitively the answer is no because here, we consider countable union and intersection rather than finite union and intersection. But, I can't think of a theorem supporting this claim.

$\\{0,0,\cdots,0\\} \times \Omega$ (where there are $n$ zeros) is in $\mathscr{A}$ for each $n$ but the intersection of these is not in$\mathscr{A}$. For unions just take complements of these sets.