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This is a follow up on a comment to one of my previous questions. What's the definition of $\omega$?

Are the following equivalent definition of $\omega$:

  1. $\omega$ is the initial ordinal of $\aleph_0$.

  2. $\omega$ is the least/first infinite ordinal.

  3. $\omega$ is the set of all finite ordinals.

  4. $\omega$ is the first non-zero limit ordinal

If yes, are there any more equivalent definitions, not on this list?

Community
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Rudy the Reindeer
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4 Answers4

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$\omega$ is defined to be the set of all finite ordinals.

This is provably equivalent to the assertion "the least infinite ordinal" or "the least limit ordinal" (note that $0$ is not a limit ordinal. It is $0$).

It can be stated as the smallest inductive set. Or the set of all ordinals whose rank is finite.

Indeed it is also defined to be $\aleph_0$.

EuYu
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Asaf Karagila
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  • What about point 1 on the list? – Rudy the Reindeer Nov 04 '12 at 10:47
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    @Matt: I’m not even sure what you mean by point 1. What is your definition of $\aleph_0$ for that point? – Brian M. Scott Nov 04 '12 at 17:41
  • @BrianM.Scott I guess I want to define it as $\aleph_0 = |\mathbb N| = |\omega|$ so that my point 1 becomes circular. – Rudy the Reindeer Nov 05 '12 at 10:25
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    Minus one for promoting falsehood in boldface. But of course, I would't down vote you or anyone else who is being nice and is trying to help me : ) – Rudy the Reindeer Nov 05 '12 at 13:43
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    @Matt: No... zero was never a limit ordinal. Soon you will tell me that the empty set has a minimal element too. – Asaf Karagila Nov 05 '12 at 13:45
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    One can define limit ordinal to be any ordinal that is not a successor ordinal. Clearly, zero is not a successor ordinal. – Rudy the Reindeer Nov 05 '12 at 13:46
  • As for the OP: I might delete this and repost it in a new thread now that my actual question has become clearer. – Rudy the Reindeer Nov 05 '12 at 13:48
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    @Matt: One can also define limit ordinals as ordinals which stand on their hands while clapping like a unicorn. That doesn't mean this is a good definition. – Asaf Karagila Nov 05 '12 at 13:51
  • @AsafKaragila In my experience, most instances of being a "limit" ordinal can be elegantly replaced with the simpler condition of not being a successor ordinal. nLab also takes the convention of defining "limit ordinal" as "any ordinal which is not a successor of any other ordinal." – user76284 Apr 07 '20 at 09:02
  • @user76284: And then $0$ is a limit ordinal. Which some people take as the definition, sure, but I prefer to separate $0$ to its own special category. – Asaf Karagila Apr 07 '20 at 09:19
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Here is a definition that works even without the axiom of infinity, in which case $\omega$ can be a proper class. Namely, $\omega$ is the class of finite ordinals. An ordinal $\alpha$ is finite if $\alpha=0=\emptyset$ or $\alpha$ is a successor ordinal that has only $0$ and other succesor ordinals as predecessors.

Michael Greinecker
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  • Before I up vote I will have to be sure that $\omega$ in your definition is actually a proper class. As of now I think it is a set. – Rudy the Reindeer Nov 05 '12 at 15:43
  • @Matt: It can be. $V_\omega$ is a model of $\mathbf{ZF}-\mathbf{Inf}+\lnot\mathbf{Inf}$ in which $\omega$ is a proper class: Michael’s definition works fine, but $\omega\notin V_\omega$. – Brian M. Scott Nov 05 '12 at 16:47
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I prefer to say: $\omega$ is the order type of the natural numbers with its usual order. All those others are theorems or common identifications.

GEdgar
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  • But if you consider a non-standard model which has non-standard integers? Then this definition fails (externally). – Asaf Karagila Nov 05 '12 at 15:06
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    @Asaf: I think your comment is irrelevant. – GEdgar Nov 05 '12 at 15:08
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    But all these definitions in the question are internal, whereas yours is external. I think it is actually quite relevant, especially in light of the other recent questions b Matt. – Asaf Karagila Nov 05 '12 at 15:30
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I'd like to summarise what I have learnt from this question:

Point (1) is circular since $\aleph_0$ is defined to be the cardinality of $\omega$.

Let's assume that we define $\omega$ to be the first ordinal of infinite cardinality. Then it must contain all finite ordinals since the ordinals are a linear order with respect to $\subseteq$. From this it is immediately clear that (2) and (3) are equivalent. It is similarly easy to see that (4) is equivalent to (3).

Rudy the Reindeer
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  • Or simply to be $\omega$. – Brian M. Scott Nov 05 '12 at 16:48
  • @BrianM.Scott Is it not ok to make the distinction between $\mathbb N$ and $\omega$ (as I described in the set theory chat)? – Rudy the Reindeer Nov 05 '12 at 20:25
  • I was talking about $\aleph_0$ there: for me it simply is $\omega$. Frankly, I tend to use $\Bbb N$ when I want to use $\omega$ but suspect that my readers won’t be familiar with it. I suppose that if I did anything with PA, I’d use it there as well. In general I don’t think of PA at all when I think of $\omega$: it’s simply the first limit ordinal. – Brian M. Scott Nov 05 '12 at 20:34
  • @Brian, Matt: There is a minor difference between $\Bbb N$ and $\omega$. Where $\Bbb N$ is the model of second-order PA, the one true model in the entire universe, $\omega$ is an ordinal, and if one works with models of set theory - and in particular with non-well founded models - then one has to make the distinctions between the various $\omega$'s. The notation of $\Bbb N$ is preserved for the unique model of second-order PA, whereas $\omega$ is just an ordinal with cool properties (measurable, supercompact, etc.) – Asaf Karagila Nov 05 '12 at 22:11