There's no cardinal $\kappa$ such that $2^\kappa = \aleph_0$

Question

I am trying to prove that there is no cardinal $\kappa$ such that $2^\kappa = \aleph_0$ .

My attempt: We suppose it exists. Since $\kappa<2^\kappa$, in particular, $\kappa<\aleph_0$. But that implies that $\kappa$ is finite. And therefore, $2^\kappa$ is finite. That leads to a contradiction.

It doesn't seem to be right, but I don't know how to proceed. Can someone help me?

Answer 1

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 $2^n$.