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

Question

What is the first cardinal number which is greater than continuum? We denote it by? Thanks very much.

Answer 1

$\mathfrak{c}^+$ 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 $\mathfrak{c} \leq \aleph_\alpha$.

Answer 2

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.

Answer 3

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.