What is aleph-one to the power of aleph-null

Question

What is $\aleph_1^{\aleph_0}$?

Can anyone shed some light. I'm not sure if this is provable or even has a value. Thanks

Answer 1

Since $\aleph_1\leq 2^{\aleph_0}$, it follows that $$2^{\aleph_0} \leq \aleph_1^{\aleph_0} \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0},$$ and hence we conclude that $\aleph_1^{\aleph_0} = 2^{\aleph_0}$.

More generally, for any infinite cardinal $\kappa$, $2^{\kappa}=\lambda^{\kappa}$ for all cardinals $\lambda$ with $2\leq \lambda\leq \kappa^+$. See here. Since $\aleph_1=(\aleph_0)^+$, the result also follows from that observation.

Note that some of the assertions above may involve or require the Axiom of Choice.