Show that power set is a sigma-algebra

**ecnanif** · November 20th 2010, 05:55 AM

Hi

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

Solution (?)

Since $\Omega$ is finite, $\exists$ $n \in \mathbb{N}$ such that $card(\Omega) \leq n$ . From set theory we know that the number of subsets of a finite set of cardinality $n$ is $2^{n}$ . Therefore,
$card(2^{\Omega}) \leq 2^{n} < \infty$ , since $n < \infty$ . Hence, $2^{\Omega}$ is finite.
$\left\{\emptyset\right\}$ and $\left\{\Omega\right\}$ is in $2^{\Omega}$ . Since $2^{\Omega}$ contains all subsets of $\Omega$ it must also contain all necessary unions, intersections and complements to make $2^{\Omega}$ a $\sigma$ -algebra.

Comments/improvements?

Thanks!

**FernandoRevilla** · November 20th 2010, 06:06 AM

Originally Posted by ecnanif

$\left\{\emptyset\right\}$ and $\left\{\Omega\right\}$ is in $2^{\Omega}$

You should say $\emptyset,\;\Omega$ instead of $\left\{{\emptyset}\right\},\;\left\{{\Omega}\right \}$ .

Regards.

Fernando Revilla

**ecnanif** · November 20th 2010, 06:08 AM

Originally Posted by FernandoRevilla

You should say $\emptyset,\;\Omega$ instead of $\left\{{\emptyset}\right\},\;\left\{{\Omega}\right \}$ .

Regards.

Fernando Revilla

Ok, but besides from this, correct?

**FernandoRevilla** · November 20th 2010, 06:10 AM

Originally Posted by ecnanif

Ok, but besides from this, correct?

Yes.

Regards.

Fernando Revilla

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