Hi

Let $\displaystyle \Omega $ be a finite set. Show that the set of all subsets of $\displaystyle \Omega $, $\displaystyle 2^{\Omega}$, is also finite and that it is a $\displaystyle \sigma$-algebra.

Solution (?)

Since $\displaystyle \Omega $ is finite, $\displaystyle \exists $ $\displaystyle n \in \mathbb{N} $ such that $\displaystyle card(\Omega) \leq n $. From set theory we know that the number of subsets of a finite set of cardinality $\displaystyle n $ is $\displaystyle 2^{n}$. Therefore,

$\displaystyle card(2^{\Omega}) \leq 2^{n} < \infty $, since $\displaystyle n < \infty $. Hence, $\displaystyle 2^{\Omega} $ is finite.

$\displaystyle \left\{\emptyset\right\}$ and $\displaystyle \left\{\Omega\right\}$ is in $\displaystyle 2^{\Omega}$. Since $\displaystyle 2^{\Omega}$ contains all subsets of $\displaystyle \Omega $ it must also contain all necessary unions, intersections and complements to make $\displaystyle 2^{\Omega}$ a $\displaystyle \sigma$-algebra.

Comments/improvements?

Thanks!