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I quote from the Wikipedia article:

"So (assuming the axiom of choice) we identify $\omega_\alpha$ with $\aleph_\alpha$, except that the notation $\aleph_\alpha$ is used for writing cardinals, and $\omega_\alpha$ for writing ordinals. "

Let's remind ourselves of the definitions:

(Def.1) The $\aleph_\alpha$ numbers are defined recursively, with $\aleph_0 = |\mathbb N| = \omega = \omega_0$ and $\aleph_{\alpha + 1} =$ the least ordinal such that it has strictly greater cardinality than $\aleph_\alpha$ for $\alpha \in \mathbf{ON}$.

(Def.2) The initial ordinal of a cardinal $\kappa$ is defined to be the least ordinal of cardinality $\kappa$. Then $\omega_\alpha$ are the transfinite initial ordinals.

Then by definition, every $\aleph$ is an ordinal and also by definition, $\aleph_\alpha = \omega_\alpha$, with or without AC. What am I missing? Thanks for clarification.

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It is perhaps good to add a link to current revision of the Wikipedia article - it might change (or be totally removed). –  Martin Sleziak Nov 3 '12 at 9:55
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2 Answers

up vote 1 down vote accepted

Perhaps they're alluding to the fact that without the axiom of choice, not all cardinals are alephs; with it, the finite ordinals and the $\aleph_\alpha$ take care of all the cardinals.

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18 seconds... :-) –  Asaf Karagila Nov 3 '12 at 8:44
    
Just to confirm: you are saying we don't need AC to get $\aleph_\alpha = \omega_\alpha$? –  Matt N. Nov 3 '12 at 8:58
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You're correct about that, Matt. –  Cameron Buie Nov 3 '12 at 9:08
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The $\aleph$ numbers are defined as the initial ordinals, without any appeal to the axiom of choice. This is done by a transfinite recursion over the ordinals. $\aleph_\alpha$ is therefore the ordinal $\omega_\alpha$, which is the unique initial ordinal such that the set of all initial ordinals strictly smaller than itself has order type $\alpha$.

In some places define cardinals only as $\aleph$ numbers, and then you need AC to assert that every set has a cardinal number. Of course you can do it without AC but then there are non-$\aleph$ cardinals as well.


One should note that before Dana Scott suggested the definition of a cardinal without the axiom of choice much later than von Neumann suggested using the $\aleph$ numbers. Historically, if so, it was "obvious" how to assign cardinals when the axiom of choice was present, but not without it.

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That's not what I asked. The passage is saying that we need AC to make the identification $\aleph_\alpha = \omega_\alpha$ and my question is, do we? (Note that this identification is not the definition of $\aleph_\alpha$). –  Matt N. Nov 3 '12 at 8:46
    
@Matt: And if you read what I wrote closely you will see that I answered that in the first line of this answer. –  Asaf Karagila Nov 3 '12 at 8:47
    
I'm sorry, but I still don't understand how this is true. –  Matt N. Nov 3 '12 at 9:33
    
@Matt: Actually the identification is the definition. –  Asaf Karagila Nov 3 '12 at 9:42
    
Is there a reason for unaccepting? –  Asaf Karagila Nov 4 '12 at 11:43
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