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

general-topologymeasure-theoryprobability theory

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:

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

  2. 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?

Best Answer

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.

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.

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