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Assuming that the GCH is false, then could there possibly exist an $n \in \mathbb N$ such that $\beth_n > \aleph_\omega$?

This would mean that the space between aleph numbers decreases more and more such that $\lim_{n \to \infty} \aleph_n = \aleph_\omega$ converges to a number on the "beth number number line" ($\aleph_\omega < \beth_\omega$).

Asaf Karagila
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abcd8849
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  • Every beth number is an aleph number. – bof Apr 09 '17 at 05:15
  • @bof How do we know that? – abcd8849 Apr 09 '17 at 05:16
  • Every infinite cardinal is an aleph – Hayden Apr 09 '17 at 05:17
  • Is there a proof of that? How do we know that the set of infinite cardinals is countable? – abcd8849 Apr 09 '17 at 05:18
  • Of course $\beth_1\gt\aleph_\omega$ is possible. – bof Apr 09 '17 at 05:19
  • Reading between the lines, I get the impression that you think the alephs end with $\aleph_\omega,$ is that right? Actually, they go on and on, there are way more than just "uncountably many" of them. – bof Apr 09 '17 at 05:21
  • Yes, you're right. My question should actually be "is there a beth number larger than all $\aleph_n, n \in \mathbb R$" – abcd8849 Apr 09 '17 at 05:25
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    For example, the answer to this question: http://math.stackexchange.com/questions/115689/can-the-cardinality-of-continuum-exceed-all-aleph-numbers-in-zf?rq=1 shows that it's impossible for $\beth_1 > \aleph_\omega$, but is it possible for $\beth_2 > \aleph_\omega$, $\beth_3 > \aleph_\omega$, ...? – abcd8849 Apr 09 '17 at 05:31
  • You cannot use the real numbers to index the $\aleph$s. http://math.stackexchange.com/questions/215977/real-cardinality-say-aleph-pi – Asaf Karagila Apr 09 '17 at 05:33
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    Also, that's not what the question you linked shows. – Asaf Karagila Apr 09 '17 at 05:35
  • "How do we know that the set of infinite cardinals is countable?" Huh?? Surely we DON'T know such a thing. Surely even just the ℶs themselves are not countable. – Wd Fusroy Oct 25 '19 at 16:31

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The answer is negative.

For example, if $\sf GCH$ holds below $\aleph_\omega$, with the exception that $2^{\aleph_0}=\aleph_2$, e.g. in Cohen's model of $\lnot\sf CH$, then $\beth_\omega=\aleph_\omega$, and $\beth_n=\aleph_{n+1}$ for $n>0$.

Moreover, given any function $f\colon\omega\to\omega$ which describes a consistent behavior of the continuum function below $\aleph_\omega$, it is consistent $f$ describes the continuum function below $\aleph_\omega$ and that $\beth_\omega=\aleph_\omega$.

At the same time, it is consistent that $\beth_1>\aleph_\omega$, so of course we cannot prove that $\beth_\omega=\aleph_\omega$ from $\sf ZFC$.

In terms of "space", the space between $\aleph$ numbers is considered to be constant, whereas the $\beth$ numbers can be spaced out further. The reason is that we can always increase $\beth$ numbers without collapsing cardinals. This is impossible for $\aleph$ numbers.

So to sum up, then, it is consistent that $\aleph_\omega=\beth_\omega$ along with many different behaviors of the continuum functions below $\aleph_\omega$ (and thus different $\beth$ assignments). And it is also consistent that $\beth_1>\aleph_\omega$.

Let me also remark that both $\aleph$ and $\beth$ numbers are indexed in terms of ordinals, and they certainly continue well after $\omega$. There is a proper class of $\aleph$ and $\beth$ numbers.

Asaf Karagila
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