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The minimality of $\omega$ relates to the fact that it's a subset of every inductive set. This answer from Magidin thoroughly explains a lot of facts about $\omega$. But, how can we be sure $\omega$ contains only $\emptyset$ and the successor of every of its elements? I mean, is there any counter examples?

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2 Answers2

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Good question.

Surprisingly, we can't. In particular, the claim that "the elements of $\omega$ are precisely: the emptyset, and all its successors" isn't first order definable.

That being said, half of this claim is first-order definable, as long as we allow an infinite list of sentences.

  1. $\emptyset \in \omega$.
  2. $\emptyset^+ \in \omega$
  3. $\emptyset^{++} \in \omega$
  4. etc.

Furthermore, we can prove each of the above sentences from the axioms of ZFC (actually, technically we prove their translations into the language of $\{\in\},$ but spiritually this isn't that important of a distinction). So $\omega$ certainly includes all the natural numbers. Also, we can prove all the distinctness theorems you might expect, like $\emptyset^+ \neq \emptyset^{++}$.

So $\omega$ is certainly an infinite set incorporating all the natural numbers.

What we can't prove using ZFC, or even formulate in its language, is the converse. If you're interested in why this is the case, I'd be happy to add some more information.

goblin GONE
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  • I'm interested certainly, though I might not be able to grasp it if it deals with more elementary logics... –  Jan 18 '14 at 03:15
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    @Fred, okay, I might have a go at writing something a bit later. No promises though, since I'm not quite sure what needs to be said. – goblin GONE Jan 18 '14 at 03:18
  • Is it the same as saying that we cannot prove that $\omega$ contains only $\emptyset$ and its successors? –  Oct 06 '15 at 02:08
  • @Fred that's right. – goblin GONE Oct 06 '15 at 03:37
  • I thought of using the fact that $(\omega , <)$ is a linear order to infer that if $\omega$ contained any extra elements they should correspond to sets containing all the elements of $\omega$, maybe simply $\omega$ itself and its successors. And as such we could use the axiom schema of specification to remove the inductive sets so that only the "common" naturals would remain, coming to demonstrate the existence of a smaller inductive set. –  Oct 06 '15 at 09:38
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To add to @user18921's answer, we can show that there are structures inside of, say, ZFC that there are structures that ZFC sees as models of the natural numbers that contain extra elements (even uncountably many). There are some neat proofs from compactness, if you feel like learning some rudimentary model theory. What this means is, provided we've formulated the problem correctly in ZFC, that there are a host of properties that first order theories can't distinguish about their models. There are "standard"/minimal models of $\mathbb{N}$ relative to ZFC (or whatever set theory)--models that are as minimal as they get in that theory--but our set theory is subject to the same ambiguity as our models of $\mathbb{N}$.

Happily, the soundness of first order logic means we can't derive anything that's actually wrong about $\mathbb{N}$ (assuming everything's consistent) even if the underlying set is non-standard. Likewise, if ZFC tells us that there are non-standard versions of $\mathbb{N}$, and we've properly captured the relation between theories and their models in ZFC, then we can have reasonable confidence in the result even if we have no clear way of knowing how non-standard the set theoretic universe gets.

Malice Vidrine
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