GHC claims that:
2^($\aleph_0 $) = $\aleph_1 \ $
2^($\aleph_1 $) = $\aleph_2 \ $ and so on.
Since the aleph function defined as:
$\aleph_0$ = $\omega$
2^($\aleph_\alpha$) = $\aleph_\alpha^+$ and so on.
Since the beth function defined as:
$\beth_0$=$\aleph_0$
2^($\beth_0 $) = $\beth_1 \ $
2^($\beth_1 $) = $\beth_2 \ $ and so on.
If we assume GCH then aleph numbers and beth numbers would not be same?
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3$\begingroup$ Yes. ${{{{}}}}$ $\endgroup$– Andrés E. CaicedoMar 1, 2017 at 17:17
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1 Answer
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You have a mistake in the definition of the $\aleph$ cardinals. $2^{\aleph_\alpha}$ is not $\aleph_\alpha^+$, at least not in general. This is $\sf GCH$, and not the definition of the $\aleph$ numbers.
The $\aleph$ numbers are defined as follows: - $\aleph_0=\omega$; - $\aleph_{\alpha+1}$ is the least ordinal whose cardinality is larger than $\aleph_\alpha$ (such ordinal exists due to Hartogs' theorem); and - If $\alpha$ is a limit ordinal, then $\aleph_\alpha=\sup\{\aleph_\beta\mid\beta<\alpha\}$.
The $\beth$ numbers, on the other hand, are defined using power sets for successor steps.
Since $\sf GCH$ is often formulated as $2^{\aleph_\alpha}=\aleph_{\alpha+1}$, we can now prove by transfinite induction that $\aleph_\alpha=\beth_\alpha$ for all $\alpha$.
answered Mar 1, 2017 at 19:37
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$\begingroup$ You have of course right. In the script of my lecture there are several definitions and theorems before the aleph sequences. I did not state them all. On the 40th page of the script (math.uni-bonn.de/ag/logik/teaching/2016WS/Aktuelles_Skript.pdf) you can see that it is stated similar to your statement with the previous Definition. (I mean the Def. 105 and then Def. 107) I see if we assume GCH, then the answer is obvious but I wanted to be sure that I do not skip any important step. $\endgroup$– munatMar 2, 2017 at 11:11