Exercise 1.1

Proof. We claim any infinite $\sigma$-algebra must be uncountable. For suppose not, and $S={\{A_i\}}_{i=1}^{\infty }$ is our countable $\sigma$-algebra on a set $X$. Then, for each $x\in X$, let $B_x:={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{x\in A_i}{A}_i$; then $B_x\in S$. We first claim that if $B_x\cap B_y\ne \varnothing$, then $B_x=B_y$. For suppose $z\in B_x\cap B_y$. Then, $B_z\subset B_x$, and if $x\notin B_z$, then $B_x\setminus B_z⊊B_x$, contradicting the definition of $B_x$. Thus, $B_x=B_z=B_y$.

Now consider ${\{B_x\}}_{x\in X}$. Then, $X={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in X}{B}_x$, and if there are only finitely many distinct $B_x$, then $S$ is also finite, a contradiction. So, there are infinitely many $B_x$, and the cardinality of $S$ must be at least $2^N$. □