Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

$\omega$ 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 $\aleph_0$.

share|cite|improve this answer
What about point 1 on the list? – Rudy the Reindeer Nov 4 '12 at 10:47
@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 4 '12 at 17:41
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 5 '12 at 13:43
@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 5 '12 at 13:45
@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 5 '12 at 13:51

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.

share|cite|improve this answer
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 5 '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 5 '12 at 16:47

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.

share|cite|improve this answer
But if you consider a non-standard model which has non-standard integers? Then this definition fails (externally). – Asaf Karagila Nov 5 '12 at 15:06
@Asaf: I think your comment is irrelevant. – GEdgar Nov 5 '12 at 15:08
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 5 '12 at 15:30
up vote 0 down vote accepted

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).

share|cite|improve this answer
Or simply to be $\omega$. – Brian M. Scott Nov 5 '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 5 '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 5 '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 5 '12 at 22:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.