# Natural numbers in Set Theory

We seem to accept the fact $(\omega,+,\times,<,0,1)^{V}$, where $V:=x=x$ is the set theoretic universe, properly reflects what is intuitively understood to be the set of natural numbers, i.e. we seem to accept that given any element $n\in{\omega}$, there is some meta-mathematical finite number $m$ (which properly reflects what it means to be finite) s.t. $n$ is obtained as the $m^{\text{th}}$ successor of $0$. Why is this the case?

I realize that you cannot state the above requirement in a coherent way using sentences in set theory. But we do seem to accept the above fact as a given. I asked a similar question as a part of an other question Meta Theory when studying Set Theory The answer given there; "we often take the universe to have the same integers as the metatheory" by which I assumed that it was an implicit assumption that $(\omega,+,\times,<,0,1)^{V}$ actually reflects natural numbers in the above, makes sense. But my question is: Why are we so sure that this is the case?

• I don't think that the expression "meta-mathematical finite number" makes sense; are you saying : "a number"? In the sense that on object in the intended model of (e.g.) first-order Peano Arithmetics ? – Mauro ALLEGRANZA Jan 22 '14 at 14:00
• Not quite. Think of it this way: A hundred apples makes sense. You just count a hundred apples. So the hundred here can be thought of as the meta-mathematical finite number. Thus informally, we can talk about what it means to be finite. My question, in a sense is: Why are we sure that $\omega$ consists only of the set theoretic counterparts for these and nothing else? – UserB1234 Jan 22 '14 at 14:10
• Well, how is $\omega$ defined for you? One way I've seen it defined is as the closure of $\{\emptyset\}$ under the successor operation, in which case it's trivial to show what you want. Other definitions may make it trickier, but it just depends. – Cameron Buie Jan 22 '14 at 14:20
• Wouldn't saying that $\omega$ is the closure make it an external definition (your concept of closure would be defined in the meta-theory not ZFC)? I agree that it forces what I want but I was hoping that it would be an internal definition. Say $\beta$, an ordinal, is a naturaL number iff for any $\alpha\leq{\beta}$ is either a zero or a successor. Now let $\omega$ be the set of all naturaL numbers. How would you show the result then? – UserB1234 Jan 22 '14 at 15:01
• It is still not clear to me (but it is my fault) if : (i) you are pointing to a "philosophical" problem : how I can reconcile our "intuitive" grasping of finite numbers with their "construction" into $ZF$ (see Paul Benacerraf's paper : What Numbers Could Not Be (1965)), or (ii) to a model theoretic issue : how may I select from the different structure satisying PA axioms, the intended one. – Mauro ALLEGRANZA Jan 22 '14 at 15:46

We understand $$\omega$$ as a model for the natural numbers because it satisfies the categorical second-order theory of $$\sf PA$$. So if we understand "the natural numbers" as the unique (up to isomorphism) model of $$\sf PA_2$$, this means that indeed $$\omega$$ is this model.

So working internally to $$V$$, this has to be the case. However, it is possible to have a model of $$\sf ZFC$$ which disagrees on its integers with its meta-theory. Still, internally, $$\sf ZFC$$ proves the induction schema, so even if $$V$$ has non-standard integers, it still thinks they can be generated by applying the successor to $$\varnothing$$ finitely many times.

This means that when $$V$$ and the meta-theory disagree on the notion of finite we have a bit of a trouble when translating induction on formulas in the meta-theory, to induction on the integers internally. So we assume this is not the case. Why can we even assume that? Well, we can't really assume that. Even if our meta-theory is $$\sf ZFC$$, the existence of a model which agrees with the universe of the integers is a stronger assumption than just assuming there is a model of $$\sf ZFC$$.

But set theory, and mathematics in general, is a utilitarian science. If the assumption is useful, and it's not "horrible" then we are likely to assume it. Since we often work with much stronger things (like transitive models, large cardinals, etc.) assuming that the universe and its meta-theory agree on the integers is not a big deal.

It should be noted that often we don't need that, and we don't care for that. If we just fix some universe of set theory in order to develop classical analysis, say, then we can forget about the meta-theory completely and just prove everything internally to the model. The need for this interaction comes up mostly when we talk about set theory itself.

And that we do, as I said, often under much stronger assumptions in consistency strength. If you have large cardinals in the universe, and the meta theory doesn't agree with the universe somehow, work internally to the universe, using some model which agrees with the universe on the integers (and much more if you want). So suddenly, this assumption is not that horrible. It's quite innocent. So we make it.

(As promised, a summary on how to prove that $$\omega$$ satisfies $$\sf PA_2$$.)

In order to say that $$\omega$$ satisfies $$\sf PA_2$$ we need to be able and say that given a statement in the language of second-order arithmetic, whether or not it is true in $$\omega$$. However in the universe of set theory, talking about subsets and predicates over $$\omega$$ is all very tangible.

So we need to encode the language, for this we need symbols for first- and second-order variables (so we fix two disjoint countable sets for that) and we fix symbols for addition, multiplication, $$0$$, $$1$$ and $$\leq$$ (we may as well use the actual sets which are these objects, but it doesn't matter).

The rules for forming a formula are the same as in first-order logic, only we are allowed to quantify over second-order variable as well, and write $$x\in A$$ when $$x$$ is a first-order variable and $$A$$ is a second-order variable.

An assignment now is a function assigning first-order variables natural numbers, and second-order variables sets of natural numbers. Since we are talking about full semantics, we are allowed to use any set in the universe.

If $$\sigma$$ is an assignment and $$\varphi(x_1,\ldots,x_n,A_1,\ldots,A_k)$$ is a formula (where $$x_i$$ are first-order and $$A_j$$ are second-order variables), then we say that $$\omega\models_\sigma\varphi(\ldots)$$ by recursion exactly as we define the satisfaction for first-order logic.

1. If $$\varphi$$ is atomic, then it is of the form $$x\in A$$ or some term $$x+y\leq z$$ or something similar to that. It is true if $$\sigma(x)\in\sigma(A)$$, etc.

2. If $$\varphi$$ is a conjunction/disjunction/material implication/negation, we simply apply the truth function on the smaller formula, using the induction hypothesis.

3. If $$\varphi$$ is $$\forall x\psi$$ where $$x$$ is a first-order variable, then $$\omega\models_\sigma\varphi$$ if and only if for every $$n\in\omega$$, $$\omega\models_{\sigma[x/n]}\psi$$ where we assign $$n$$ to the now-free variable $$x$$; similarly for $$\exists x\psi$$.

If $$\varphi$$ is $$\forall A\psi$$ where $$A$$ is a second-order variable, then $$\omega\models_\sigma\varphi$$ if and only if for every $$M\subseteq\omega$$, $$\omega\models_{\sigma[A/M]}\psi$$ where we assign the second-order variable the value $$M$$. Similarly for existential quantifiers.

All in all, this is exactly what we think it should be. No funky business going on. I should remark that there are variants of second-order logic where we take a limited range of subsets. For example Henkin semantics only allows us to assign to second-order variables sets which are first-order definable. This significantly weakens the logic. But here we talk about full semantics, so everything is a fair game.

Now from the induction theorem in $$\sf ZF$$ we have, in fact, a formula $$\sf Sat_2$$ which takes in an encoding of a formula, and an assignment and returns whether or not the formula is true or false in $$\omega$$.

So now, given that $$\sf PA_2$$ is in fact finite, we can ask whether or not the conjunction of all the axioms hold in $$\omega$$. It is not hard to verify all the first-order axioms are true; the second-order induction axiom is true simply because the definition of "inductive set" is the same in the context of $$\sf PA_2$$ in $$\omega$$ and in the full universe (since they agree on the successor and $$0$$).

Since $$\sf ZF$$ proves that $$\omega$$ is in fact the smallest inductive set, it follows that every subset of $$\omega$$ which is inductive, is in fact $$\omega$$ itself. So the second-order induction axiom holds as well. Therefore $$\omega\models\sf PA_2$$, as wanted.

• Thank you. Could you please point me to a reference for the facts you claim about viewing $\omega$ as the model for the second order theory of PA? That is the part I'm having trouble with. As I said with regards to the previous question; I'm not doubting anything about model theory and the existence of non-standard, non-transitive or non-$\omega$ models. I was just confused as to why in $w^{V}$ was considered to reflect the natural numbers. – UserB1234 Jan 22 '14 at 16:29
• Danul, it's easy to verify that the arithmetic axioms hold (i.e. the interpretation of the arithmetic of ordinals is the same), once you define the satisfaction relation on $\omega$ (i.e. given an arbitrary formula, of any logic expressible in set theory - e.g. second-order logic - whether or not it is true or false in $\omega$), it is easy to show that the induction schema is provable from the induction theorem of $\sf ZF$. – Asaf Karagila Jan 22 '14 at 16:33
• I have to head out soon, but when I get home I will take some time to write this up into my answer. – Asaf Karagila Jan 22 '14 at 16:34
• Thank you. I don't have much experience with second order logic. I would like to learn a little bit about it. So if you have an idea of where I should start, it would be very helpful. – UserB1234 Jan 22 '14 at 16:38
• For second-order logic, the canonical text is Stewart Shapiro's Foundations without Foundationalism: a case for second-order logic. This in the Oxford Logic Guides series from OUP, and should be in any decent library. – Peter Smith Jan 22 '14 at 17:32

The question asks why we assume that the natural numbers of the metatheory are the natural numbers of some model of ZFC. Of course there are nonstandard models of ZFC, so we can only hope that some "standard" models will have this property. In general, a model that has the same natural numbers as the metatheory is called an $\omega$-model.

First, three caveats:

1. Nothing that we can prove within ZFC can justify this. It would be naively possible that ZFC is consistent but not $\omega$-consistent (and thus has no $\omega$-model), and in that case we could still prove all the same things in ZFC.

2. We have to firmly distinguish between the $\omega$ of the metatheory and the $\omega$ of each model of ZFC. Unfortunately, there is no established notation to keep these very clear.

3. Much of the answer depends on how one views the metatheory.

There are at least two justifications for thinking ZFC has an $\omega$-model.

1. There is a famous interpretation of ZFC in terms of the cumulative hierarchy. This interpretation claims that if the cumulative hierarchy is developed in the metatheory, it will satisfy ZFC. Thus, this interpretation suggests that ZFC has an $\omega$-model.

2. Optimism. Set theorists feel as if they know ZFC relatively well, and that ZFC captures the "real" notion of set. This is a pragmatic argument, of course.

On the other hand:

1. We know that the statement "ZFC is consistent" does not imply "ZFC has an $\omega$-model" within ZFC. So we cannot hope to use the mere consistency of ZFC to prove within ZFC that it has an $\omega$-model.

2. Recent work of Joel David Hamkins, colleagues, and others on multiverse set theory shows that is it also perfectly consistent to assume in a multiverse metatheory that ZFC has no $\omega$-model at all. The show that, if ZFC is consistent, then it is consistent that there is a multiverse of models of set theory in which every model is non-well-founded relative to some other model. In a multiverse like that, there is no $\omega$-model of set theory, since $\omega$ (assuming it it well defined, in this context) has to be well founded.

• I don't know if this quite makes sense: but my question was more along the lines of: Why do we believe the meta-theoretic natural numbers are the natural numbers of the set theoretic universe? My thinking was along these lines: suppose you take a platonistic view. Then there is an absolute universe of set theory, in the sense every set is there in that universe. The study of set theory (and as a result everything studied in set theory, which is pretty much everything) seems to assume that the "actual $\omega$" in the actual universe is a true reflection of the natural numbers. – UserB1234 Jan 23 '14 at 14:57
• This idea seems ok if you regard it in the following way. Assuming you can see the universe of Set Theory and assuming that your meta theory is strong enough to handle infinitistic arguments; if the $\omega$ in the set theoretic universe had a non-standard element then you should be able to somehow make a correspondence between the meta-theoretic natural numbers and the ones in the set theoretic universe. This is very similar to what you do with non-standard models, in the sense someone living in the non-standard model doesn't think it's non-standard but someone who lives in the – UserB1234 Jan 23 '14 at 15:15
• set-theoretic universe knows it is non-standard because they have more information. I have two issues with this though: Relating things in the meta theory and the object theory is always tricky business, so without a solid math to back it up I'm reluctant to admit my intuitive argument. The second is, the natural numbers in the meta theory had to come from somewhere. If the meta theory is also ZFC, then since the meta theoretic natural numbers also come from there and we are assuming that they are exactly what we want them to be, we might as well assume that the set theoretic universe has – UserB1234 Jan 23 '14 at 15:23
• the correct integers. In fact it makes zero sense not to do so. In which case it kind of boils down to a matter of belief. I was just wondering whether there is something here that I'm missing. – UserB1234 Jan 23 '14 at 15:24