# Is a boolean algebra closed under countable disjunction/conjunction?

I'm just curious if the properties in a sigma algebra is also satisfied in a boolean algebra. In a boolean algebra, the two operators are closed under finite operations, but can we say they are closed under countably infinite operations?

• You might be interested in complete boolean algebras, see here. On this page is also the example of the boolean algebra consisting of all subsets of an infinite set which are finite or have finite complement. This algebra is not complete because taking an infinite union might lead to a set not contained in this algebra. – Matthias Klupsch May 11 '16 at 9:27

Not in general, no. You can see this wiki-page for some counter-examples. For example, the boolean algebra of finite and co-finite sets of integers is not complete, for let $E$ be the set of all singletons of even integers, and assume $P$ is an upper bound for $E$. Then clearly $P$ is infinite, and hence it must be co-finite. Since it is co-finite, it must contain a singleton of an odd integer (for otherwise its complement would also be infinite), so if we remove this odd integer, we still have an upper bound, and this upper bound is below $P$. Hence $E$ has no supremum.
A boolean algebra is called complete if it is closed under arbitrary suprema and infima, and it is called a boolean $\sigma$-algebra if it is closed under countable suprema and infima. So $\sigma$-algebras are really just a special case of boolean algebras.