With the cardinality of the natural numbers as $|\mathbb{N}| = \aleph_0$ and its powerset as $|\mathcal{P}(\mathbb{N})| = 2^{\aleph_0}$, the continuum hypothesis and the axiom of choice says that there is no set such that

$$ \aleph_0 < |S| < 2^{\aleph_0} = \aleph_1 $$

The whole hierarchy of aleph numbers can be constructed from $\aleph_n = \mathcal{P}(\aleph_{n-1})$. Each aleph is constructed from powerset iterations of the natural numbers. Is there any non-finite set $S$ where this is not true?

$$ |\aleph_0| < |S| \notin \{\aleph_0, \aleph_1, \aleph_2, \ldots \} $$

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    $\begingroup$ Sure. Take the union over all natural numbers of the $\aleph_n$. That's new. Then take the powerset and the game starts again. $\endgroup$ Feb 6, 2014 at 19:39
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    $\begingroup$ $\aleph_{n+1}$ is the cardinal successor of $\aleph_n$. It is consistent with $\mathsf{ZFC}$ that $\aleph_{n+1} < 2^{\aleph_n} = | \mathcal{P} ( \aleph_n ) |$ for all $n$. (Heck, its consistent that $\aleph_n < 2^{\aleph_0}$ for all $n$.) $\endgroup$
    – user642796
    Feb 6, 2014 at 19:42
  • $\begingroup$ Reinforcing Arthur's comment: Your understanding of the notation is wrong! $\aleph_1$ is not defined as $2^{\aleph_0}$, it is defined as the smallest cardinal that is greater than $\aleph_0$. The continuum hypothesis is that the two definitions are equal. $\endgroup$
    – TonyK
    Feb 6, 2014 at 19:49

1 Answer 1


Some (valid) points.

  1. $\aleph_{n+1}$ is NOT defined as $2^{\aleph_n}$. $\aleph_{n+1}$ is the least cardinality of an ordinal which is strictly larger than $\aleph_n$.

  2. The continuum hypothesis, if so, is formulated as the assumption $2^{\aleph_0}=\aleph_1$. If it is false, then there is an intermediate set of cardinality $\aleph_1$ (assuming the axiom of choice, anyway). It was shown that given any cardinal, the continuum is consistently larger. So we can't even decide using the usual axioms of set theory, how many cardinals are between the real numbers and the natural numbers.

  3. We can prove, however, that some $\aleph$ numbers cannot be equal to the cardinality of the continuum. For example the first cardinal which is larger than infinitely many [infinite] cardinals, denoted by $\aleph_\omega$, cannot be the cardinality of the continuum.

  4. There is, however, another hierarchy, of $\beth$ [read: Beth] numbers ($\aleph$ is the first letter of the Hebrew alphabet, and $\beth$ is the second letter). It is defined as $\beth_0=\aleph_0$ and $\beth_{\alpha+1}=2^{\beth_\alpha}$ (I'm omitting the limit stages, for now).

    In terms of $\beth$ numbers we can in fact prove that $2^{\aleph_0}=\beth_1$, and the continuum hypothesis can now be stated as $\aleph_1=\beth_1$.

  5. Now your question can be rewritten as either of two questions. If we change the occurrences of $\aleph$ to $\beth$, then every $\beth$ number is defined as a power set of the previous one (and I am omitting the limit case, again); and so the answer is "no, there isn't any", or we can what are the $\aleph$ numbers which are smaller than $\beth_1$, but as I pointed out, we can show that this is consistently any possible value $0$.

  • $\begingroup$ Thank you for the informative answer, I think it clears up a few mistakes in both my nomenclature and my understanding of the ordinals. I was trying to lookup the Hebrew "B" transfinite number and couldn't simply google for the letter: to help those who follow, they are called Beth numbers. $\endgroup$
    – Hooked
    Feb 6, 2014 at 21:01

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