How to simplify the following:
$2^{\aleph_0}(\aleph_0+\aleph_0)^{2^{\aleph_0}}$
Thank you for every help.
How to simplify the following:
$2^{\aleph_0}(\aleph_0+\aleph_0)^{2^{\aleph_0}}$
Thank you for every help.
Use that
(i) For cardinals $\kappa$ and $\lambda$ with $\kappa \le \lambda$ and $\lambda$ infinite you have that $\kappa + \lambda = \lambda$ and hence $\aleph_0 + \aleph_0 = \aleph_0$
(ii) For cardinals $\kappa \le \lambda$ where $\lambda$ is infinite you have $\kappa \cdot \lambda = \lambda$
(iii) For an infinite cardinal $\lambda$ and $2 \le \kappa \le \lambda$ you have $\kappa^\lambda = 2^\lambda$
Then
$2^{\aleph_0}(\aleph_0+\aleph_0)^{2^{\aleph_0}} \stackrel{(i)}{=} 2^{\aleph_0}\aleph_0^{2^{\aleph_0}} \stackrel{(ii)}{=} \aleph_0^{2^{\aleph_0}} \stackrel{(iii)}{=} 2^{2^{\aleph_0}}$
Hint: Prove that $(\aleph_0)^{2^{\aleph_0}}=2^{2^{\aleph_0}}$.