It is stated that by Zermelo’s theorem, every cardinal is an aleph. But what is the difference between cardinals and alephs? I thought that alephs were just a way to denote cardinals (just for notation), but then this theory doesn't make sense.

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    $\begingroup$ cardinals wear red caps, alephs wear black yarmulkahs. $\endgroup$ – sds Apr 13 '17 at 22:11
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    $\begingroup$ Probably it might be useful where it is stated. In general it is good to provide context for the question. (Even though in this specific case you received good answers even without adding the source of the claim.) $\endgroup$ – Martin Sleziak Apr 14 '17 at 10:06
  • $\begingroup$ For example, if you include in your post: "Something like this appeared in the lecture I am attending, see here", then somebody could be able to point out how Alephs were defined in one of the previous lectures and that Zermelo's theorem is related to well-orderings. $\endgroup$ – Martin Sleziak Apr 14 '17 at 10:12
  • $\begingroup$ I do not know whether this is really a course you are attending (I have just tried to google for formulation similar to the one in your question), but if it was, the above information might be useful for you. And link to the course materials could be useful to answerers, since they would see what you have already been taught. $\endgroup$ – Martin Sleziak Apr 14 '17 at 10:13

The issue is the axiom of choice.

The $\aleph$ numbers are the cardinalities of well-ordered sets: if $A$ can be well-ordered, then there is some least ordinal $\alpha$ which can be bijected with $A$, and this is the cardinality of $\alpha$ (and such ordinals in general are called "initial ordinals"). In case $A$ is infinite, we get an $\aleph$ number (and there's really nothing interesting to say about the cardinalities of finite sets).

But if the axiom of choice fails, not every set can be well-ordered! And so if we want to speak of the cardinality of a non-well-orderable set, we need to use something other than $\aleph$s.

At this point it's worth saying a few words about what cardinality is.

First up, we have the "equinumerosity" relation $\equiv$: we write "$A\equiv B$" if there is a bijection between $A$ and $B$. This is easy to define, and there's no problem with it if the axiom of choice fails.

Now what's the cardinality of a set $A$? Well, here's the idea: we want to associate some object $\vert A\vert$ to every set $A$, such that $\vert A\vert=\vert B\vert$ iff $A\equiv B$ (that is, $\vert A\vert$ is an $\equiv$-invariant: if you know what $\vert A\vert$ is, then you know what $A$ is equinumerous with). One natural choice (this one is due to Frege) is to look at the entire $\equiv$-class itself - e.g. the cardinal "$2$" is just the collection of all $2$-element sets. Unfortunately, this is a proper class, so this doesn't work well with ZFC.

Instead, we have to be a little ad hoc. The natural way to fix Frege's idea is via Scott's trick: we let $\vert A\vert$ be the set of all sets equinumerous with $A$ and of minimal rank, and this is indeed a set (and we can think of it as an "initial segment" of the class Frege cares about). This definition, again, works independently of the axiom of choice (although it if we drop both choice and foundation, and in fact I think there's no good way to define cardinality in the absence of both axioms - instead, you have to work with the relation "$\equiv$" alone).

Now if $A$ is well-ordered, we can do better: as observed above, we can pick out a specific set which is equinumerous with $A$! And that's the $\aleph$ number of $A$. In the presence of choice, there's no reason to use the Frege-style definition above, and we simply equate "cardinality" with "$\aleph$-number". But if choice fails, we can't find canonical representatives to measure the size of some sets, so we have to do something more involved, like Scott's trick (and note that at the linked question there is some argument for the Scott approach actually being more natural, which I have some sympathy with).

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    $\begingroup$ There is very much that is really interesting about the cardinalities of finite sets. Normally though, we think of it as being about numbers, not cardinals. $\endgroup$ – Paul Sinclair Apr 13 '17 at 23:09
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    $\begingroup$ @PaulSinclair Indeed. For example, one such interesting result about cardinalities of finite sets is this: If $N,A,B,C$ are non-empty finite sets, with $A.B$ disjoint, then the cardinality of the set of maps from $N$ to $A$ union the set of maps from $N$ to $B$ can only equal the cardinality of the set of maps from $N$ to $C$ if there exists an injective map $N\to\{\emptyset,\{\emptyset\}\}$. $\endgroup$ – Hagen von Eitzen Apr 14 '17 at 11:09
  • $\begingroup$ @HagenvonEitzen - indeed. One must be quite wiley to prove that. $\endgroup$ – Paul Sinclair Apr 14 '17 at 16:34
  • $\begingroup$ @HagenvonEitzen 20 minutes it took me. Evil person.:D Very cool formulation though. $\endgroup$ – DRF Apr 14 '17 at 17:29

Well. The finite ordinals are also cardinals, but they are not $\aleph$ numbers.

But without the axiom of choice, there can be sets which cannot be well-ordered, and therefore their cardinality cannot be "measured" using $\aleph$ numbers. In that case we fallback to Scott cardinals, which allow us to define cardinals even in the absence of choice, for any set.

  • $\begingroup$ Argh, good point re: finite sets - forgot that in my answer! $\endgroup$ – Noah Schweber Apr 13 '17 at 18:37
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    $\begingroup$ To quote Nelson Muntz, Ha ha! :P $\endgroup$ – Asaf Karagila Apr 13 '17 at 18:37
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    $\begingroup$ +1, and I approve of Simpsons references. $\endgroup$ – Noah Schweber Apr 13 '17 at 18:57
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    $\begingroup$ Alessandro Andretta gave a talk in Arctic Set Theory 2017 that used that package, by the way. $\endgroup$ – Asaf Karagila Apr 13 '17 at 18:59

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