# Number of elements in a finite $\sigma$-algebra is even

To prove the number of elements in a $$\sigma$$-algebra is even, is it enough if I argue that for every $$A\in\mathcal{F}, A^c\in\mathcal{F}$$, where $$\mathcal{F}$$ is a $$\sigma$$-algebra, and thus the elements have to occur in pairs? Is there a flaw with this argument?

• The claim is true so long as $\mathcal{F}$ is a $\sigma$-algebra on a nonempty set. Otherwise $\emptyset=\emptyset^\mathsf{c}$ (as $\emptyset\setminus \emptyset=\emptyset$). – Hayden Apr 26 '20 at 17:49
• @Hayden yes, the underlying set is non-empty by assumption. thanks! – learner Apr 26 '20 at 17:55

It is enough. Suposse you have $$2n+1$$ elements, then by the pigeonhole principle you can choose $$n$$ elements $$\{x_1, \dots\ x_n\}$$ such that $$x_i^c \neq x_j$$ for all $$i,j \leq n$$. This implies that that for the remaining $$n+1$$ elements $$n$$ of those are of the form $$x_i^c$$, so one element doesn't have an complementar in the algebra.
Let $$F$$ be a finite $$\sigma$$-algebra on a set $$A\ne\emptyset.$$ For $$x\in F$$ let $$g(x)=\{x,A\setminus x\}.$$ Every $$g(x)$$ has exactly $$2$$ members, and if $$x,y\in F$$ then either $$g(x)=g(y)$$ or $$g(x)\cap g(y)=\emptyset.$$ So $$\{g(x):x\in F\}$$ is a $$partition$$ of $$F$$ into pairwise-disjoint $$2$$-member subsets.