There's no cardinal $\kappa$ such that $2^\kappa = \aleph_0$
Your proof is fine.Since Aleph-0 is defined to be the leasr infnite cardinal,anythung less is finite.And if n is finite,so is $2^n$.
There's no cardinal $\kappa$ such that $2^\kappa = \aleph_0$
Your proof is fine.Since Aleph-0 is defined to be the leasr infnite cardinal,anythung less is finite.And if n is finite,so is $2^n$.