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The first Aleph number is $\aleph_0$, and my question is this: why is the second Aleph number defined to be $\aleph_1 = 2^{\aleph_0}$? If I remember correctly, it had something to do with power sets and the fact that the cardinality of a power set is always greater than the cardinality of the original set. (Also, if I'm not mistaken, $\aleph_1 = 2^{\aleph_0}$ is true only if the continuum hypothesis is true. Is this the case?)

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  • $\begingroup$ They aren't. You're confusing them with Beth numbers, which are defined this way: $\beth_{\alpha + 1} = 2^{\beth_{\alpha}}$. $\endgroup$ – BrianO Jan 2 '16 at 0:35
  • $\begingroup$ $\aleph_\omega$ is not of the form $2^x$. $\endgroup$ – bof Feb 10 at 12:37
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$\aleph_1$ is not defined to be $2^{\aleph_0}$, it's defined to be the "next" cardinal number following $\aleph_0$. As you said, that equality holds is stated by the continuum hypothesis; and as we know from Gödel and Cohen, this can't be proved from the ZFC axioms.

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  • $\begingroup$ Is there a simple way to explain why the continuum hypothesis states that the next cardinal number following Aleph-null is 2^Aleph-null? $\endgroup$ – user265554 Jan 1 '16 at 11:02
  • $\begingroup$ One way to explain the motiviation behind the continuum hypothesis is that the cardinality of the set of natural numbers is $\aleph_0$, and it is easy to prove that the cardinality of the set of real numbers, often called $\mathfrak c$, is $2^{\aleph_0}$. If $\mathfrak c$ weren't the next cardinal number, then we'd have subsets of the reals which were neither countable ($\aleph_0$) nor of the same size as $\mathbb R$. $\endgroup$ – Frunobulax Jan 1 '16 at 11:06
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    $\begingroup$ @user265554 What do you mean by "why?" The historical reason is that for a long time many (most?) mathematicians believed that $\aleph_1=2^{\aleph_0}$ was true from first principles. It wasn't until 1963 that Paul Cohen proved it was independent. On the other hand there is some reason to think of it as the "minimal" choice since CH holds in the constructible universe $L$. $\endgroup$ – DRF Jan 1 '16 at 11:08
  • $\begingroup$ How does one prove that the cardinality of the set of real numbers is 2^Aleph-null? $\endgroup$ – user265554 Jan 1 '16 at 11:11
  • $\begingroup$ No need anymore, thank you all! $\endgroup$ – user265554 Jan 1 '16 at 11:17
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Basically, given any cardinal $\kappa$, we know that $2^{\kappa}> \kappa$. This is known as Cantor's theorem, which basically states that there is no surjective function form a set $A$ to its powerset $\mathcal{P}(A)$.

Furthermore, we do not know for any infinite cardinal $\kappa$ whether there exists a cardinal $\lambda$ such that $2^{\kappa} > \lambda > \kappa$. The generalized continuum hypothesis states that there does not exist such $\kappa$.

The aleph numbers are defined as

  • $\aleph_0$ is the least infinite cardinal.
  • $\aleph_{\alpha+1}$ is the least cardinal larger than $\aleph_{\alpha}$.
  • $\aleph_{\alpha}=\sup\{\aleph_{\lambda}: \lambda<\alpha\}$ for all limit ordinals $\alpha$.

The beth numbers are defined as

  • $\beth_0$ is the least infinite cardinal.
  • $\beth_{\alpha+1}=2^{\beth_{\alpha}}$ and
  • $\beth_{\alpha}=\sup\{\beth_{\lambda}: \lambda<\alpha\}$ for all limit ordinals $\alpha$.

Note that by Cantor's theorem, $\beth_{\alpha+1}>\beth_{\alpha}$. Hence we may show that $\beth_{\alpha}\geq\aleph_{\alpha}$, and we may reformulate the continuum hypothesis as $\beth_1=\aleph_1$ and the generalized continuum hypothesis as $\beth_{\alpha}=\aleph_{\alpha}$ for all ordinals $\alpha$.

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