# Suppose GCH, then Aleph function and Beth function are isomorphic?

GHC claims that:
2^($\aleph_0$) = $\aleph_1 \$
2^($\aleph_1$) = $\aleph_2 \$ and so on.
Since the aleph function defined as:
$\aleph_0$ = $\omega$
2^($\aleph_\alpha$) = $\aleph_\alpha^+$ and so on.
Since the beth function defined as:
$\beth_0$=$\aleph_0$
2^($\beth_0$) = $\beth_1 \$
2^($\beth_1$) = $\beth_2 \$ and so on.
If we assume GCH then aleph numbers and beth numbers would not be same?

• Yes. ${{{{}}}}$ Mar 1, 2017 at 17:17

You have a mistake in the definition of the $\aleph$ cardinals. $2^{\aleph_\alpha}$ is not $\aleph_\alpha^+$, at least not in general. This is $\sf GCH$, and not the definition of the $\aleph$ numbers.
The $\aleph$ numbers are defined as follows: - $\aleph_0=\omega$; - $\aleph_{\alpha+1}$ is the least ordinal whose cardinality is larger than $\aleph_\alpha$ (such ordinal exists due to Hartogs' theorem); and - If $\alpha$ is a limit ordinal, then $\aleph_\alpha=\sup\{\aleph_\beta\mid\beta<\alpha\}$.
The $\beth$ numbers, on the other hand, are defined using power sets for successor steps.
Since $\sf GCH$ is often formulated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$, we can now prove by transfinite induction that $\aleph_\alpha=\beth_\alpha$ for all $\alpha$.