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

As in the title: Why does $\omega$ have the same cardinality in every (transitive) model of ZF? I've been thinking about this for some time now.

Can someone show me how to show this by showing me a proof? Thanks a lot.

share|cite|improve this question
Matt, what exactly is there to prove? $\omega$ is, by definition, the first nonzero limit ordinal. – Andrés E. Caicedo Nov 3 '12 at 21:44
@Andres: of course, by compactness there are models $X$, $Y$ of set theory with $|\omega^X| = \aleph_1$ and $|\omega^Y| = \aleph_2$. But such models won't be transitive. I think this is where the question is going. – Carl Mummert Nov 3 '12 at 21:59
@CarlMummert Sure, ok, that makes sense. – Andrés E. Caicedo Nov 3 '12 at 22:11
@AndresCaicedo Actually, how would that make it countable? (I assumed its definition was that it's the initial ordinal of $\aleph_0$, that is, $\omega = \omega_0$. Perhaps that is incorrect?) – Rudy the Reindeer Nov 4 '12 at 7:42
I wonder who voted to close this question. I think it's actually a good question. – Asaf Karagila Nov 5 '12 at 13:03
up vote 6 down vote accepted

Suppose $X$ and $Y$ are transitive models of set theory. Then the elements of $X$ and $Y$ are sets, and the set membership relations $\in^X$ and $\in^Y$ are the normal set membership relation $\in$.

Hence we can build up the following sequence of results, using the assumption of transitivity:

  • $\emptyset^X$ (the unique element $z$ of $X$ such that $X \models (\forall y) [y \not \in z]$) and $\emptyset^Y$ are both the real empty set, $\emptyset$.

  • So $1^X = \{\emptyset^X\}$ and $1^Y$ are both $1 = \{ \emptyset\}$.

  • Similarly, for each finite ordinal $n$, $n^X = n^Y = n$.

  • Moreover, for any $z \in X$, if $X$ satisfies the sentence "z is a finite ordinal", then $z$ really is a finite ordinal. The same holds for $Y$.

  • Thus $\omega^X = \omega = \omega^Y$. So not only do $\omega^X$ and $\omega^Y$ have the same cardinality, they are literally the same set.

share|cite|improve this answer
Dear @CarlMummert, thank you for this clear and concise answer. The reason I asked the question was because someone had told me that $\omega$ had the same cardinality in every model of ZF because we could "write it down as a formula". Is this related to what you wrote in your answer? – Rudy the Reindeer Nov 4 '12 at 7:28
(I think it could be false because it would also apply to non-transitive models.) – Rudy the Reindeer Nov 4 '12 at 7:29
@Matt N.: I cannot say exactly what they might have meant, but they might have been referring to some of the general results about absoluteness. One place this is developed carefully and slowly is in Kunen's textbook; he spends quite a while developing facts about absoluteness that are needed for forcing, and this fact about $\omega$ is one of them. Once someone is familiar with the general results they will think it is very obvious that $\omega$ is the same in every transitive model. – Carl Mummert Nov 4 '12 at 12:17
Thank you for pointing me to Kunen. – Rudy the Reindeer Nov 5 '12 at 13:38

If a model is transitive then its ordinals are real ordinals. In particular it means that any finite ordinal in the model is actually finite, and therefore $\omega$ is the real $\omega$.

It is important to note, however, that there is a possibility that countable sets are not countable in the model; and that it is possible for non-transitive (and non-well founded) models to exist in which there is an uncountable number of finite ordinals, of course as Carl points out, internally the finite ordinals are always a countable set. It is only when we look at the model from an external point of view we can tell it has an uncountable number of finite sets.

The important point to notice is that we often assume that there is an "absolute" universe in which we do mathematics, and this universe is a model of ZFC. While this model is not a set, when we talk about models of ZFC we talk about set models. Now there is an internal and external issues to discuss. It is always the case with set models that there is an element of the set-model which the set model "sees" as having a very big cardinality, but the universe sees as having a "small" cardinality.

For example, if $M$ is a countable transitive model then the real numbers of $M$ is a countable set, but not internally.

For non-transitive models the natural numbers could be uncountable as well, externally, at least. However for a transitive model this is impossible.

share|cite|improve this answer
Of course you know this, but it may help to point out that any model $X$ of ZFC satisfies the sentence "the set of finite ordinals is countable"; but when we look at the model externally we may see that the set in the model which it thinks is the set of finite ordinals is actually uncountable. – Carl Mummert Nov 3 '12 at 22:01
I understand how a set can be countable in the real universe and uncountable in some model it is part of, with the $\aleph_1$ of a countable model the standard example. This is the first I had heard that a set could be countable in a model and uncountable in the real universe-aren't all sets in the model (including the bijection with $\aleph_0$ of this set) in the real universe? – Ross Millikan Nov 3 '12 at 23:40
@Ross Millikan: take a nonstandard model of ZFC $M$ so that the set that $M$ believes is $\mathbb{N}$ has cardinality $\aleph_1$. Of course $M$ thinks its own $\mathbb{N}$ is countable. A model of this form can be made with the compactness theorem - but models like this will not be transitive, nor well-founded. The bijection in $M$ between $\mathbb{N}^M$ and itself is in the real world, but the real world doesn't think this bijection shows anything about countability. – Carl Mummert Nov 4 '12 at 3:27

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.