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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?

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

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  • $\begingroup$ I feel this answer complicates the proof. The proof in question and first comment is so simple and crisp $\endgroup$ – Nagabhushan S N Apr 26 '20 at 18:13
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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.

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