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$\aleph_0$, $\aleph_1, \aleph_2$ and so on are indexed by a natural number so shouldn't there be countably many infinities?

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    $\begingroup$ Those aren't the only infinities... $\endgroup$ – Thomas Andrews Oct 6 '13 at 23:39
  • $\begingroup$ I'm pretty sure they'res an infinite number of infinities? $\endgroup$ – Rivasa Oct 6 '13 at 23:41
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After you exhaust all the $\aleph_n$, you still have $\aleph_\omega$ which is a cardinality larger than all the $\aleph_n$'s. Then you have $\aleph_{\omega+n}$, for integer $n$, and so on.

And you have many many many more cardinals. In fact, for every ordinal $\alpha$ you have cardinal $\aleph_\alpha$. Since there are uncountable ordinals, there are at least uncountably many cardinals. But in fact, the collection of cardinals does not make a set, it is a proper class.

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  • $\begingroup$ Is there an easy way to see that there is an $\aleph_\alpha$ for each ordinal $\alpha$? $\endgroup$ – celtschk Oct 18 '14 at 10:15
  • $\begingroup$ Sure, by definition of the $\aleph$ numbers. $\aleph_0=\omega_0$; $\aleph_{\alpha+1}$ is the least ordinal which is not of cardinality $\leq\aleph_\alpha$ (often denoted by $\omega_\alpha$) and if $\delta$ is a limit then $\aleph_\delta=\sup_{\beta<\delta}\aleph_\delta$. $\endgroup$ – Asaf Karagila Oct 18 '14 at 10:35
  • $\begingroup$ But I don't see how you can get more than countably many cardinalities by that procedure. $\endgroup$ – celtschk Oct 18 '14 at 14:54
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    $\begingroup$ Let's do this backwards, what's bothering you in this argument? $\endgroup$ – Asaf Karagila Oct 18 '14 at 14:59
  • $\begingroup$ Let's start with $\aleph_0$. Now with the $\alpha+1$ rule, I can generate $\aleph_1$, $\aleph_2$ and so on. Basically, everything $\aleph_n$, $n\in\mathbb N$. With the limit rule I then get $\aleph_\omega$. Then I get $\aleph_{\omega+1}$ etc., then again with the limit rule $\aleph_{2\omega}$, and indeed $\aleph_{m\omega+n}$. I even see how you can extend that to arbitrary polynomials of $\omega$. But those are still countably many ordinals, and thus countably many cardinals associated to them. So, how do you get beyond countability? $\endgroup$ – celtschk Oct 18 '14 at 15:06

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