Possible inaccuracy in Wikipedia article about initial ordinals

Question

"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.

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 03 '12 at 09:55

Cameron Buie · Accepted Answer · 2018-04-19T11:24:25.477

4

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

edited Apr 19 '18 at 11:24

answered Nov 03 '12 at 08:41

Cameron Buie

102,994

3

18 seconds... :-) – Asaf Karagila Nov 03 '12 at 08:44
Just to confirm: you are saying we don't need AC to get $\aleph_\alpha = \omega_\alpha$? – Rudy the Reindeer Nov 03 '12 at 08:58
1

You're correct about that, Matt. – Cameron Buie Nov 03 '12 at 09:08

Asaf Karagila · Answer 2 · 2012-11-03T09:36:46.420

3

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.

edited Nov 03 '12 at 09:36

answered Nov 03 '12 at 08:41

Asaf Karagila

393,674

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 $\aleph_\alpha$). – Rudy the Reindeer Nov 03 '12 at 08:46
@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 03 '12 at 08:47
I'm sorry, but I still don't understand how this is true. – Rudy the Reindeer Nov 03 '12 at 09:33
@Matt: Actually the identification is the definition. – Asaf Karagila Nov 03 '12 at 09:42
Is there a reason for unaccepting? – Asaf Karagila Nov 04 '12 at 11:43
I had first accepted Cameron's answer, then accepted yours because after the third edit or so it did answer my question quite well. But then I thought that that's not a good reason to unaccept Cameron because he answered my question in a comment, before yours did. – Rudy the Reindeer Nov 04 '12 at 18:50

Possible inaccuracy in Wikipedia article about initial ordinals

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