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In his popular book Set Theory: An Introduction to Independence Proofs, Kunen gives the following definitions on the bottom of page 145:

Let $\mathcal{A} = \lbrace A, E \rbrace$ be a structure for the language of set theory. Let also $\mathcal{A} \models ZF$. We call $\mathcal{A}$ an $\omega$-model iff there is no $a \in A$ such that $\mathcal{A} \models “a \in \omega”$ but $a \neq n^{\mathcal{A}}$ for each $n$.

He then proceeds with the following assertion:

If $\mathcal{A} \models ZF$, then for each formula $\phi$ in the metatheory, there is a corresponding $\phi^{\mathcal{A}} \in A$, where $\phi^{\mathcal{A}}$ is the interpretation of $\ulcorner\phi\urcorner$ in $\mathcal{A}$ (where $\ulcorner\phi\urcorner$ is a constant symbol—usually an element of $\omega^{< \omega}$—meant to represent $\phi$ in the language). If $\mathcal{A}$ is an $\omega$-model, then these are the only formulas of $\mathcal{A}$, but if $\mathcal{A}$ is not an $\omega$-model, then $\mathcal{A}$ has non-standard formulas whose lengths are infinitely large natural numbers.

Basically, I am trying to make sense of the statement in bold. First of all, is Kunen claiming that a non-$\omega$-model may contain non-standard formulas or that it will necessarily contain such formulas? If so, how can we reach that conclusion? It seems to me that even if $A$ has non-standard elements, we still have no way of knowing if $\phi^{\mathcal{A}}$ is or isn't a standard natural number of $A$, regardless of what formula $\phi$ we start with.

What am I missing? Some compactness argument perhaps?

Bonus question: What is "a length of size equal to an infinitely large natural number" in this context? I mean it's one think to talk about non-standard elements of a model, and a completely different one to associate these elements with "size" in the metatheory. How do these non-standard formulas look-like?

gen-ℤ ready to perish
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Pellenthor
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    It seems like your main issue is that you do not have a definition of what "formula of $\mathcal{A}$" means. Without such a definition you cannot possibly say anything about such formulas. – Eric Wofsey Apr 13 '20 at 14:51

2 Answers2

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For every natural $n$, $\phi_n$ is a sentence, where $\phi_0$ is $\forall x\,(x=x)$ and $\phi_{n+1}$ is $(\phi_n\land\phi_n)$. By recursion, there is a sentence in the theory that codes this claim and so, for any model, for any $n$ that, from the point of view of the model, is a natural number, there is an object of the model that the model interprets as the sentence $\phi_n$. This holds even if $n$ is nonstandard.

Of course, if $n$ is nonstandard, this object $\phi_n$ is not really a formula, but the model cannot see that.

Andrés E. Caicedo
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  • This is very useful thanks. Just to make sure I fully understand this: If $a$ is non-standard, would that make $\phi_a$ be "$a$ copies" of $\phi_0$? – Pellenthor Apr 14 '20 at 11:50
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    Yes, from the point of view of the model, that's what $\phi_a$ is. – Andrés E. Caicedo Apr 14 '20 at 11:56
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    Pellenthor, $2^{\alpha}$ copies, actually. (That's gonna be WEERD!) Note that it doubles up each iteration. If you want $\alpha+1$ copies, take $\phi_{n+1} = \phi_n \wedge \phi_0$, which just appends one $\phi_0$ each increment instead of doubling up the string each increment. (If you want $\alpha$ copies, rename $\phi_0$ to $\phi_1$ :) ) – The_Sympathizer Apr 16 '20 at 00:18
  • Good catch @The_Sympathizer! I was indeed considering the "linear" case for some reason. Perhaps in an attempt to not miss the forest for the trees... :) But you are of course correct! – Pellenthor Apr 16 '20 at 12:40
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The whole "point", one might say, of an $\omega$-model is that its natural numbers only consist of the "standard" natural numbers. Since pretty much by definition, any model of ZFC must contain a set that it would "call" as "$\mathbb{N}$", we can inquire as to the contents of this set and whether or not they are only the "standard" natural numbers or, whether or not that they also include nonstandard numbers. $\omega$-models' "$\mathbb{N}$"s only include standard natural numbers.

So if we are not in an $\omega$-model, then that means that the model's "$\mathbb{N}$" must contain some non-standard numbers. Where that translates to non-standard formulas is that "formulas" are also an object we can formulate within the set theory and thus also can undergo "promotion" via the transfer principle. To see this, note that (as just one of an endless number of possible ways), we can encode a formula as a specific kind of function from a natural number to $\{ 0, 1 \}$ or better, in purely set-theoretic terms, to $\{\emptyset, \{\emptyset\}\}$, where the interpretation of such a function is that it indexes the bits of the formula when its graphic symbols are encoded in some kind of binary-based encoding, say, something like ASCII or UNICODE, and then taking that to be a string of binary bits (0 or 1).

But note now: because we have non-standard numbers, we can now have some formula-like objects which are functions with domain a nonstandard number. Such things are formulas of nonstandard length. Moreover, if it did not contain such formulas, that would mean it would have naturals, that it would recognize as such, and yet that would not be able to map out to $\{0, 1\}$ in ways that ZFC says can happen, and so such a model would fail to be a model of ZFC.

Finally, what does such a formula "look" like, as like a visualization? Well, imagine an infinitely long trail of logical symbols like you'd usually think, e.g.

$$\neg(A \vee [B \wedge C] \wedge \neg(\neg A) \vee \cdots$$

trailing off forever, but then also, somewhere "out there in the hazy mist of the foggy borderland between the definitively finite and definitively infinite", you can dream off to other strigns of symbols...

$$\cdots \vee A \vee A \vee A \vee [\neg A] \wedge B \wedge \cdots$$

where it continues now bidirectionally on both sides and, just as how a nonstandard natural looks, there's a dense linecloud-collection of these doubly-open-ended infinite chains. However, the model, just as it cannot see the nonstandard naturals are nonstandard, also can't see that this strange thing is not a formula. The symbols (or better, the bits in the encoding) will be indexed purely by nonstandard numbers, e.g. $\neg$ above will be located at, say, $(\mbox{some infinitely big 'root'}) - 6000^{\mathrm{googolplex}^\mathrm{moser}}$. And, of course, the formula must be generable by some procedure that could be carried out in ordinary ZFC, and extended to the nonstandard length.

The_Sympathizer
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  • Thank you, this is what I was looking for. So is the "key" that if $A$ contains a nonstandard number $a$, then it will also interpret $\omega^{< \omega}$ differently than an $\omega$-model? By that I mean that now $\mathcal{A}$ "thinks" that the function $a \rightarrow \omega$ is just another finite sequence of strings and thus it must correspond to yet another "normal" formula $\phi$. In reality, outside of $\mathcal{A}$, $a$ is some infinite ordinal, thus preventing $\phi$ from being a formula. Like ordinals, the larger $a$ gets, the "weirder" $\phi$ gets. Am I thinking about this right? – Pellenthor Apr 14 '20 at 11:38
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    @Pellenthor : Absolutely - these same considerations apply to ordinal numbers, too. Non-$\omega$ ZFC models will also contain "non-standard ordinals" that are strictly between all the standard finite ordinals and $\omega$, and have the same order structure as the non-standard "$\mathbb{N}$". And this will also mean the class of ordinals will not be well-ordered "from outside", but it will "think" it is "from inside" and so its "idea" of what a well-order is will also differ from what yours may be. – The_Sympathizer Apr 16 '20 at 00:04
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    Though I should caution that with one point I just saw: $a$ will not be an infinite ordinal outside the model $\mathcal{A}$. It will be some non-archimedean ill-ordered number. This is because the set of "ordinals" in the model will not be externally well-ordered any more than $\mathbb{N}$ is. The "ordinals" will still satisfy the initial-segment property that if $\beta$ is an $\mathcal{A}$-ordinal with $\beta < \alpha$ then $\beta \subset \alpha$, but won't be externally well-ordered under this ordering. – The_Sympathizer Apr 16 '20 at 00:16
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    The tension is solved because we can extrospect that there are some initial segments which are not supped by any $\beta$, such as the standard naturals themselves, but $\mathcal{A}$ cannot introspect those: it does not feature them as sets to begin with. There is a ${"\mathbb{N}"} \in \mathcal{A}$, but $\mathbb{N} \notin \mathcal{A}$. – The_Sympathizer Apr 16 '20 at 00:16
  • This is very interesting @The_Sympathizer, thank you! I was also concerned myself, because you mentioned "bidirectionality" which is not something ordinals satisfy. Intuitively, it sounds like - and please correct me if I am wrong - we can apply a simple compactness argument to show that there is an infinite decreasing sequence above $\mathbb{N}$, to "violate" well-orderness from the outside. Incidentally, would you happen to have any study references to non-archimedean ill-ordered numbers that goes a bit deeper on the concept? – Pellenthor Apr 16 '20 at 12:32