Why does (finite) $\sigma$-algebra work in powers of 2? The cardinality of a (finite) $\sigma$-algebra is $2^k$ for some $k \in \mathbb{N}$.
There are some proofs (like this) around, but how is it claimed in the proofs that "every element of $\sigma$-algebra is the (disjoint) union of members", i.e. works in powers of 2.
 A: An explanation without (explicit) induction: let $\Sigma$ be a finite $\sigma$-algebra of subsets of $S$.  Identify each $s \in \Sigma$ with its characteristic function $\chi_s : S \rightarrow \{0,1\}$.  These $\chi_s$ are closed under addition mod 2: $\chi_s + \chi_{s'} = \chi_{s \Delta s'}$ where $\Delta$ is the symmetric difference.  Thus we have a bijection from $\Sigma$ to a vector space $V$ over the $2$-element field.  Therefore if $\Sigma$ is finite then its cardinality is $2^k$ where $k = \dim V$.  QED
A: By induction. The number of members of the Boolean algebra on a set with $0$ members is $1=2^0.$ Suppose the proposition is true for sets of at most $n$ members.Let $S$ have $n+1$ members and let $B$ be a Boolean algebra on $S.$ Choose $x\in S.$ $$\text {Let } C=\{b\in B : x\not \in B\}.\quad \text {Let } S^*=\cup C.$$ Since $B$ is finite, we have $S^*\in B.$and so $S^*\in C$. Now $C$ is a Boolean algebra on $S^*.$ (For complements in $S^*,$ if $b\in C,$ then $p\not \in S^*\backslash b=S^*\cap (S\backslash b)\in B,$ so $S^*\backslash b\in C.)$ Since $S^*$ has at most $n$ members, the number of members of $C$ is $2^k$ for some positive integer $k.$ $$\text {Now let } D= \{b\in B :p\in B\} \quad \text {and  } e=\cap D.$$ Since $B$ is finite, we have $e\in B$, and $e\in D.$ $$\text { We have } e\cap S^*=e\cap (\cup C)=\phi$$ $$\text { because } b\in C\to S\backslash b\in D\to b\cap e\subset b\cap (S\backslash b)=\phi.$$ Finally, consider the families $$T= \{b\cup e :b\in C\}\quad\text { and } U=C\cup T .$$  $$\text {We assert that } U \text { has } 2^{k+1} \text { members and that } U=B.$$ First, if $b_1,b_2 \in C$ then $$(b_1\cap e=\phi=b_2\cap e) \to  [\;(b_1\cup e =b_2\cup e)\to b_1=b_2\;].$$ Therefore $T$ has the same number of members that $C$ does, which is $2^k.$ And $T\cap C=\phi$ because $p$ belongs to every member of $T$ and to no member of $C.$ So the number of members of $U$ is $$2^k+2^k=2^{k+1}.$$ Second, we have $U\subset B,$ but for any $b\in B$ we have $$(1): p\not \in b\to b\in C\to b\in U.$$  $$(2):  p\in b\to b\supset e$$ (by the def'n of $e$); and $b\backslash e\in C$ (because $p\in e\in S$), so $$(p\in b)\to b=(b\backslash e )\cup (b\cap e)=(b\backslash e) \cup e \in U.$$  therefore $B\subset U$. We have $B\subset U \subset B$, so $U=B$. 
