What is the first cardinal number which is greater than continuum?

$\mathfrak{c}^+$ or $( 2^{\aleph_0} )^+$; the $^+$ denotes that we are taking the cardinal successor. Its value in the (transfinite) sequence of aleph numbers cannot be determined by ZFC because the value of $\mathfrak{c} = 2^{\aleph_0}$ cannot be determined by ZFC.

The Continuum Hypothesis is the conjecture that $\mathfrak{c} = \aleph_1 = \aleph_0^+$, and was proven to be independent of ZFC by Kurt Gödel (ZFC cannot prove $\mathfrak{c} \neq \aleph_1$) and Paul Cohen (ZFC cannot prove $\mathfrak{c} = \aleph_1$). Furthermore Paul Cohen's proof showed that given any aleph number $\aleph_\alpha$ ZFC cannot prove $\mathfrak{c} \leq \aleph_\alpha$.

The notation $\beth_1$ represents the cardinality of the continuum. See also, beth numbers.

Thus, by definition, $(\beth_1)^+$ is the first cardinal number which is strictly greater than $\beth_1$. See also, successor cardinal.

Whether $(\beth_1)^+ = \beth_2$ is independent of ZFC. See also, generalized continuum hypothesis.

The axioms of ZFC can prove that there exists $\alpha$ such that $\frak c=\aleph_\alpha$, but the axioms themselves are insufficient in order to prove much on that $\alpha$. We can prove that $\alpha\neq 0$ and that $\alpha$ does not have cofinality of $\omega$, whatever that might be.

But besides these two facts we can't say anything intelligible on what exactly is $\alpha$. We know it is possible to have a universe of set theory where $\alpha=1$ and another where $\alpha=2$.

Therefore we cannot really say much on what is $\frak c^+$, or $\aleph_{\alpha+1}$, since we don't know what is $\aleph_\alpha$ in this case. We can, however, prove that it exists and give it a symbol such as:

• $\left(2^{\aleph_0}\right)^+$;
• $\frak c^+$;
• $\left(\beth_1\right)^+$

And so on.