TITLE: Is $\aleph_0^{\aleph_0}$ smaller than or equal to $2^{\aleph_0}$?
QUESTION [11 upvotes]: Possible Duplicate:
What's the cardinality of all sequences with coefficients in an infinite set? 

Is $\aleph_0^{\aleph_0}$ smaller than or equal to $2^{\aleph_0}$?
I thought I saw this kind of statement somewhere, but I do not remember it.
Can anyone show me the proof of it? 
Thanks.

REPLY [11 votes]: Depending on your precise definition of "cardinal", the theorem below may require the Axiom of Choice (in fact, to make it really useful it does require it).
Theorem. For all cardinals $\lambda$ and $\kappa$, with $\kappa$ infinite, if $2\leq \lambda\leq\kappa^+$, then $\lambda^{\kappa}=2^{\kappa}$.
Proof. $2^{\kappa}\leq \lambda^{\kappa}\leq (\kappa^+)^{\kappa}\leq (2^{\kappa})^{\kappa} = 2^{\kappa\kappa} = 2^{\kappa}$. $\Box$
(AC is used to get $\kappa\kappa=\kappa$; the assertion that every infinite set $X$ is bijectable with $X\times X$ is equivalent to the Axiom of Choice).
Corollary. $\aleph_0^{\aleph_0} = 2^{\aleph_0}$.
Proof. Take $\lambda=\aleph_0$, which satisfies $2\leq \lambda \leq \aleph_0^+=\aleph_1$. $\Box$