# Comparing the two cardinals $\aleph_0^{\aleph_0}$ and $2^{\aleph_0}$ [duplicate]

Is $\aleph_0^{\aleph_0}=2^{\aleph_0}$ or $\aleph_0^{\aleph_0}>2^{\aleph_0}$ Why?

## marked as duplicate by Namaste, Cameron Buie, Asaf Karagila♦, Andrés E. Caicedo, MicahFeb 28 '13 at 4:21

• @CogitoErgoCogitoSum: I don't think anyone here will argue that cardinal arithmetic is a bit wonky, and often counterintuitive, but it's hardly fallacious. If you don't understand why $2^{\aleph_0}>\aleph_0$, I recommend you read this wonderful and intuitive answer describing why a power set is always strictly larger than the starting set. P.S.: Your username cracks me up. – Cameron Buie Feb 28 '13 at 3:59
$\aleph_0^{\aleph_0}\le\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}\le\aleph_0^{\aleph_0}$.