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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?

woterem
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    To add to the answers below, there are many operations in cardinal arithmetic that are more subtle than talking products or powers. (To mention just one example, you may want to look at the smallest size of a covering (or cofinal) family of countable subsets of $A$ rather than to $A^{\aleph_0}$.) These operations typically give you values undecided by the usual axioms, and one applied to $\aleph_0$, may very well give you the value $\omega_1$ in some models, regardless of what $2^{\aleph_0}$ is in these models. – Andrés E. Caicedo Sep 23 '12 at 21:09
  • (than taking products) – Andrés E. Caicedo Sep 28 '12 at 07:08

2 Answers2

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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:

  1. How do we know an $ \aleph_1 $ exists at all?
  2. Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
Community
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Asaf Karagila
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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.

Brian M. Scott
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