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In complex analysis, there is a function called Euler's Gamma function. Whenever given a positive integer $n+1$, it will return $n!=\prod_{i=1}^{i < n+1}i$.

I'm not sure if there is similar function for infinite cardinals such that
$$\Gamma(\aleph_\alpha)=\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa$$, but at least we can evaluate the value of that production.

So my question: Is $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$?

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    $\begingroup$ Regarding your second proof: in general $2^{\aleph_{\alpha}} \ne \beth_{\alpha}$. (But I can see what you were getting at.) $\endgroup$ Commented Apr 15, 2013 at 18:28
  • $\begingroup$ @CliveNewstead Thanks, I'm going to fix it. $\endgroup$
    – Popopo
    Commented Apr 16, 2013 at 1:43
  • $\begingroup$ You have rendered my answer useless with the two edits. $\endgroup$
    – Asaf Karagila
    Commented Apr 16, 2013 at 2:03
  • $\begingroup$ @AsafKaragila Sorry, in that time I have mistook the definition of beth numbers (mistook that $2^{\aleph_\alpha}=\beth_{\alpha+1}$). $\endgroup$
    – Popopo
    Commented Apr 16, 2013 at 2:18
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    $\begingroup$ meta.math.stackexchange.com/questions/9017/… (Note how with your latest edit you have rendered the accepted answer somewhat irrelevant just as well, it may have been better to post a new question altogether, or in case the information given to you here was sufficient to solve it on your own, leave it with the mistakes you have made.) $\endgroup$
    – Asaf Karagila
    Commented Apr 16, 2013 at 11:45

2 Answers 2

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The study of infinite products is very subtle in general. The "factorial" is reasonably well understood. In 1925, Tarski proved that $\prod_{\xi<\beta}\aleph_\xi=(\aleph_{\bigcup \beta})^{|\bigcup\beta|}$ (for details, see this blog post of mine). Since $\prod_{n\in\omega}n=2^{\aleph_0}$, we get that $$ \Gamma(\aleph_\alpha)=2^{\aleph_0}\prod_{\xi<\alpha}\aleph_\xi=2^{\aleph_0}(\aleph_{\bigcup\alpha})^{|\bigcup\alpha|}. $$ (Whether or not this is $\beth_\alpha$ is independent of $\mathsf{ZFC}$ in general, and clearly depends on the size of the two powers that appear in this displayed expression. As suggested in the comments, it may be worth pointing out that, for example, if $\alpha=\tau+1$ where $\tau$ is a countable successor ordinal and $2^{\aleph_0}<\aleph_\tau$, then $\Gamma(\aleph_\alpha)=2^{\aleph_0}\aleph_\tau^{|\tau|}=\aleph_\tau^{\aleph_0}=\aleph_\tau<2^{\aleph_\tau}\le2^{\aleph_\alpha}$.)

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  • $\begingroup$ It may be worth pointing out that if the continuum is small, then it is often the case that $\aleph_\alpha^{|\alpha|}$ is much smaller than $2^{\aleph_\alpha}$. $\endgroup$
    – Asaf Karagila
    Commented Apr 16, 2013 at 10:24
  • $\begingroup$ Thank you very much. Besides, a little tip: when $\alpha=\beta+1$ then $\Gamma(\aleph_\alpha)$ should be $2^{\aleph_0}\aleph_{\beta}^{|\beta|}$. $\endgroup$
    – Popopo
    Commented Apr 16, 2013 at 11:19
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    $\begingroup$ Hehe, yes, sorry, not all ordinals are limit ordinals. :-) Fixed now. $\endgroup$ Commented Apr 16, 2013 at 13:32
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(Note: This answer was given to a previous revision of the question)

Certainly not. Consider a model of $\sf ZFC$ such that for every countable ordinal, $2^{\aleph_\alpha}=\aleph_{\omega_1+1}=\beth_1$.

Let $\alpha$ be a countable ordinal and consider the following, $$\prod_{\beta<\alpha}\aleph_\beta\leq\prod_{\beta<\alpha}\aleph_\alpha = \aleph_\alpha^{\aleph_\alpha}=2^{\aleph_\alpha}=\beth_1\ll\beth_\alpha.$$

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