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We have $\aleph_0$, $\aleph_1$, and so on. But why wouldn't there be an $\aleph_\pi$, for example? What is the proof that the types of infinities are a countable set?

Jessica Singer
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The cardinalities are $\aleph_\alpha$ where $\alpha$ is an ordinal. So you can have $\aleph_\omega$ (which is the least cardinality larger than $\aleph_0,\aleph_1,\aleph_2,\ldots)$ or even $\aleph_{\omega_1}$ or more. So there are uncountably many, but not because they are indexed by reals, but because there is a proper class of them (i.e. too many to be a set of any size, let alone a countable one).

Francis Adams
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  • Thank you for your answer! Sorry if it's a dumb question, but couldn't there be a mapping from $(0, 1)$ to the $\aleph_\alpha$? (For example, $\aleph_1$ would map to $0.1$, $\aleph_{\aleph_2}$ would map to $0.02$ ) – Jessica Singer Jun 15 '23 at 03:26
  • If you map $\aleph_0$ to $a\in(0,1)$, where would you map $\aleph_1$? Note that this is the smallest cardinal larger than $\aleph_0$ so if you want an order isomorphism, you'd need a smallest real number larger than $a$ which doesn't exist... If you dont care about the order, you could do that (not for all cardinals but for the first $2^\omega$ many). – fweth Jun 15 '23 at 06:34
  • @JessicaSinger Like fweth said, since the ordinals are well-ordered you can't map $(0,1)$ to the ordinals in an order preserving way. But even if you wellordered $(0,1)={r_\beta:\beta<2^{\aleph_0}}$, and mapped $r_\beta$ to a cardinal $\aleph_{\alpha_\beta}$, then you still have a set of cardinals. The union is a cardinal, and then there must be a cardinal bigger than that.

    The basic idea is that if you have collection of cardinals indexed by a set, then the union is a cardinal, which has a successor. This is also why the collection of cardinals is not itself a set.

     – Francis Adams Jun 15 '23 at 12:38
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    @FrancisAdams Incidentally--and at risk of causing more confusion than it's worth--a fun fact is that in the absence of the axiom of choice, it's consistent that there is an order-embedding of $(0,1)$ into the cardinals. (The non-well-ordered cardinals, of course... not into the $\aleph$'s.) – spaceisdarkgreen Jun 16 '23 at 00:22