# Show that power set is a sigma-algebra

• Nov 20th 2010, 06:55 AM
ecnanif
Show that power set is a sigma-algebra
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.

Thanks!
• Nov 20th 2010, 07:06 AM
FernandoRevilla
Quote:

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
• Nov 20th 2010, 07:08 AM
ecnanif
Quote:

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?
• Nov 20th 2010, 07:10 AM
FernandoRevilla
Quote:

Originally Posted by ecnanif
Ok, but besides from this, correct?

Yes.

Regards.

Fernando Revilla