If a set is closed under unions and intersections is it closed under complements?

Question

I'm thinking of sigma algebras here, which are (nonempty) sets closed under countable unions, countable intersections, and complements.

But you only need 2 of these conditions to guarantee the third:

0. If a (nonempty) set is closed under countable unions and complements, then it is closed under countable intersections (countable De Morgan).

1. If a (nonempty) set is closed under countable intersections and complements, then it is closed under countable unions (countable De Morgan).

Now I ask:

• If a (nonempty) set $X$ is closed under countable unions and countable intersections, then is it closed under complements? (Does this change if $X$ is finite, countable, or uncountable?)

(I say "set" because in ZFC everything is a set, but people often call these "families" or "collections", ie. sets of sets.)

Bonus questions:

• If a set is closed under finite unions, then it is closed under countable unions?
• If a set is closed under finite intersections, then it is closed under countable intersections?

Answer 1

Let's say $X$ is a family of subsets of $\Omega$.

One problem with having unions and intersections but not complements is that maybe there's an element $x \in \Omega$ which is not found in any set in $X$. No amount of unions and intersections will get you $x$, but every complement will contain $x$.

Even if $\bigcup X = \Omega$, we have problems. For example, let $\Omega = \mathbb R$ and let $X$ consist of all intervals of the form $(-\infty, a)$ and $(-\infty, a]$. This is closed under unions and intersections, so we'll never get an interval that's infinite in the other direction.

A similar construction works for subsets of the finite set $\{1,2, \dots, n\}$ or the countably infinite set $\mathbb N$, too.

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Most topologies are examples of why the answers to your bonus questions are no. If $X$ is the set of all open sets in $\mathbb R$ (with respect to the usual topology) then $X$ is closed under arbitrary unions and finite intersections - but not under infinite intersections. For example, $\bigcap_{n=1}^\infty (-\frac1n, \frac1n) = \{0\}$, which is not open.

Similarly, the set of all closed sets will have finite unions and arbitrary intersections - but not infinite unions.

Answer 2

For any $X$, pick $Y\subsetneq X$, then $P(Y)$ is closed under arbitrary intersection/union, but not complement. In measure theory, set difference $X\setminus Y$ is more often used than complement. Pick an increasing sequence of sets $X_1\subsetneq X_2 \subsetneq X_3 \subsetneq \cdots$, then $\{X_1, \cdots, X_2, \cdots\}\cup \{\cup_i X_i\}$ is closed under arbitrary union and intersection, but not difference set.

All finite sets of an infinite set $X$ is closed under finite unions, but not countable unions.

All cofinite sets of an infinite set $X$ is closed under finite intersections, but not countable ones.