Why are Aleph numbers by definition of the form $2^x$?

Question

The first Aleph number is $\aleph_0$, and my question is this: why is the second Aleph number defined to be $\aleph_1 = 2^{\aleph_0}$? If I remember correctly, it had something to do with power sets and the fact that the cardinality of a power set is always greater than the cardinality of the original set. (Also, if I'm not mistaken, $\aleph_1 = 2^{\aleph_0}$ is true only if the continuum hypothesis is true. Is this the case?)

Answer 1

$\aleph_1$ is not defined to be $2^{\aleph_0}$, it's defined to be the "next" cardinal number following $\aleph_0$. As you said, that equality holds is stated by the continuum hypothesis; and as we know from Gödel and Cohen, this can't be proved from the ZFC axioms.

Answer 2

Basically, given any cardinal $\kappa$, we know that $2^{\kappa}> \kappa$. This is known as Cantor's theorem, which basically states that there is no surjective function form a set $A$ to its powerset $\mathcal{P}(A)$.

Furthermore, we do not know for any infinite cardinal $\kappa$ whether there exists a cardinal $\lambda$ such that $2^{\kappa} > \lambda > \kappa$. The generalized continuum hypothesis states that there does not exist such $\kappa$.

The aleph numbers are defined as

• $\aleph_0$ is the least infinite cardinal.
• $\aleph_{\alpha+1}$ is the least cardinal larger than $\aleph_{\alpha}$.
• $\aleph_{\alpha}=\sup\{\aleph_{\lambda}: \lambda<\alpha\}$ for all limit ordinals $\alpha$.

The beth numbers are defined as

• $\beth_0$ is the least infinite cardinal.
• $\beth_{\alpha+1}=2^{\beth_{\alpha}}$ and
• $\beth_{\alpha}=\sup\{\beth_{\lambda}: \lambda<\alpha\}$ for all limit ordinals $\alpha$.

Note that by Cantor's theorem, $\beth_{\alpha+1}>\beth_{\alpha}$. Hence we may show that $\beth_{\alpha}\geq\aleph_{\alpha}$, and we may reformulate the continuum hypothesis as $\beth_1=\aleph_1$ and the generalized continuum hypothesis as $\beth_{\alpha}=\aleph_{\alpha}$ for all ordinals $\alpha$.