Cardinal arithmetic and first uncountable cardinal Cardinal arithmetic does not seem to open its way to the existence of $\aleph_1$ that is not $2^{\aleph_0}$, as any operation on $\aleph_0$ would lead to $\aleph_0$ or $2^{\aleph_0}$ and $2^{2^{\aleph_0}}$ or so forth.
According to my knowledge, choice does not determine whether continuum hypothesis is right or wrong, so what is going on with this?
 A: No, cardinal arithmetic is equipped with a successor function, much like the natural numbers. The successor of a cardinal $\kappa$ is the smallest cardinal which is strictly larger than $\kappa$. The successor is denoted by $\kappa^+$.
$\aleph_1$ exists (regardless to the axiom of choice) because we can prove there exists an uncountable ordinals, and therefore there exists a least uncountable ordinal. By definition of being the least uncountable ordinals it is a cardinal (=an ordinal which is not in bijection with any smaller ordinal), and it is exactly $\aleph_1$.
You are correct however that the axiom of choice does not determine the value of $2^{\aleph_0}$, it is however needed to ensure that it is a well-ordered cardinal.

For further reading:


*

*How do we know an $ \aleph_1 $ exists at all?

*Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
A: Nothing much is going on. There are two simple successor-like operations on infinite cardinals. One is exponentiation: from $\kappa$ you get $2^\kappa$, but you can’t say much more about its size than that it’s bigger than $\kappa$. The other is the actual successor operation: from $\kappa$ you get $\kappa^+$, which is the smallest ordinal that admits no bijection onto $\kappa$. This is the real successor to $\kappa$: it’s the smallest cardinal strictly bigger than $\kappa$. The generalized continuum hypothesis is the assertion that $2^\kappa=\kappa^+$ for all infinite $\kappa$, i.e., that these two successor-like functions are really the same function; as you say, it is independent of ZFC. The axiom of choice enters the picture when you want to prove that the cardinal $2^\kappa$ is well-orderable; without AC you can’t guarantee this. The successor $\kappa^+$, in contrast, is always well-ordered.
