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I'm still fairly confused by model theory so bear with me here. The question is, if $M$ is a model of ZF (or perhaps ZFC if it makes any difference), and $M_1$ is a transitive submodel of $M$, and $x\in M_1$ is finite with respect to $M$, then is $x$ necessarily finite with respect to $M_1$?

"$x$ is finite" means that there exists a bijection between $x$ and some finite ordinal $n$.

I mostly understand why "$x$ is countable" is not absolute. But then I came across the Wikipedia article that says "$x$ is finite" is not an absolute property either.

But surely it is... right? We know that "$x$ is empty" is absolute, and "$x$ has one element" is absolute, "$x$ has two elements" is absolute, and so on. Here is my attempted by proof by induction that this holds for all $n\in\omega$:

Let us assume "$x$ has $n-1$ elements" is an absolute property, let $\emptyset\ne x\in M_1$, and suppose there exists a bijection $f:n\to X$ in $M$.

Without making any reference to $f$, let $x_0\in x$ be arbitrary and form the subset $$x\setminus \{x_0\}.$$ Clearly $x\setminus \{x_0\}\in M_1$ and it is an $(n-1)$ element set with respect to $M$. But by our inductive hypothesis, having $n-1$ elements is an absolute property, so there exists a bijection $g: (n-1)\to x\setminus\{x_0\}$ in $M_1$. But now $$g\cup (n-1,x_0)$$ is a bijection of $n$ with all of $x$.

So we're done, right? What, if anything, is wrong with that proof? Help me understand this seemingly absurd statement that finiteness is not absolute.

spaceisdarkgreen
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Alex Eustis
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  • This question is similar to: Is there an absolute notion of the infinite?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Lee Mosher Jul 23 '24 at 14:05
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    @LeeMosher The difference is that this concerns transitive models. – spaceisdarkgreen Jul 23 '24 at 15:22
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    Are you assuming $M_1$ satisfies some decent fragment of ZF? If not, things get complicated (and at a certain point the definition of finite no longer makes sense). If so, then yes, being finite is absolute for transitive models. I don't know what wikipedia is talking about... seems misleading (they are a little vague, but seem to be both assuming transitivity and being a model of ZF). – spaceisdarkgreen Jul 23 '24 at 15:25
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    @spaceisdarkgreen I meant to say that $M_1$ fully satisfies all of ZF. Thank you for your reply. All the evidence seems to indicate that finiteness is indeed absolute in this context (transitive models of ZF), and the Wikipedia page contains a mistake. – Alex Eustis Jul 23 '24 at 15:31
  • @AlexEustis Are you assuming that $M$ itself is transitive? That is the context of the Wikipedia article where indeed finiteness is absolute, but I interpreted your question as saying $M$ is an arbitrary model of ZFC and $M_1$ is a submodel which is transitive relative to $M$ (i.e. if $x\in M_1$ and $y\in_M x$ then $y\in M_1$), in which case I suspect finiteness is not absolute. – Eric Wofsey Jul 24 '24 at 03:59

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Yes, you're right. Your argument works and I think the wikipedia page is misleading at best.

(I'll phrase things in terms of the outer model $M$ being the set-theoretical universe.$^*$)

In what's maybe a bit more of a general result, if $M_1$ is a transitive class (or set) closed under union, pairing, and set difference, then it is closed under finite subsets. The proof is just a slightly simpler version of your inductive argument.

As for why this implies the result in question, if $X\in M_1$ is finite, then by the closure properties, a witnessing bijection $f:n\to X$ of $X$'s finiteness will be a finite subset of $M_1$ and thus $f\in M_1$, and the property of $f$ being such and such is of course absolute.

Actually, if you go through the argument carefully, you can see the assumption that $M_1$ is transitive or even well-founded is easily avoidable, and so the inward direction of absoluteness is actually completely general.

Where transitivity of $M_1$ is used is in going outward, and in that case, we actually only need that $M_1$ is an $\omega$-model. If we have an $X\in M_1$ and there is an $f\in M_1$ that witnesses finiteness of $X$ in $M_1,$ then $f$ actually gives a real one-to-one correspondence between a natural number $n\in \omega^{M_1}$ and $X$. So the only obstacle to outward absoluteness is that $n$ might be externally infinite in the case where $M_1$ has nonstandard $\omega.$

$^*$Eric Wofsey points out that this is a bit cavalier in terms of how you phrased the problem, as a submodel need not be definable in the original model so the internal induction argument may not work. But I think it's still probably closer to the sense meant on wikipedia (given the reference to Jech chapter 12, especially), and your argument works on this assumption, so stand by my first sentence.

spaceisdarkgreen
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    The problem is, what if $M$ has nonstandard natural numbers? Then $M_1$ may not be closed under taking subsets that $M$ thinks are finite. – Eric Wofsey Jul 23 '24 at 19:57
  • @EricWofsey If $M_1$ is a class in $M$, then $M$'s natural numbers are an initial segment $M_1$'s, right? Clearly I reverted to just thinking about $M$ as the metatheory, and I'm pretty sure this last part I added is right in that context, but I think it might still be fine in general. – spaceisdarkgreen Jul 23 '24 at 20:15
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    $M$ has to in fact have the exact same natural numbers as $M_1$, since $\omega$ is just the first ordinal that is not a successor and that is clearly absolute. The issue though is that there might be some other set which $M$ thinks is finite but $M_1$ doesn't contain its bijection to a nonstandard natural number. (This can only happen if $M_1$ is a not actually even a class from $M$'s perspective, i.e. it is not definable in $M$, since if it were definable in $M$ you can do the induction argument internally to $M$.) – Eric Wofsey Jul 24 '24 at 03:55
  • @EricWofsey Ahh I see, you're talking about absoluteness between models and submodels (in the usual sense of the term) which is indeed the way OP had phrased the question. I had been thinking in terms of absoluteness between the universe and a model, as was the case in the Jech reference given in wikipedia (and then generalizing to when the model isn't necessarily standard in what I added after the edit), and when I said "it's still fine" I meant when the universe is not necessarily externally an $\omega$-model. Are we on the same page now? – spaceisdarkgreen Jul 24 '24 at 08:38
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    Yeah, that’s probably actually the context they meant, in which case everything you’re saying makes sense. – Eric Wofsey Jul 24 '24 at 13:31
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The finiteness of $X$ has to do with the existence of an epimorphism arrow $F \to X$ from some predetermined collection of sets $\{F\}$ deemed "finite" (finite ordinals, as you say.) The existence of such arrow depends on the axioms and choice of finite ordinals.

TheAlertGerbil
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    Why? "Finite choice" is not an additional axiom, it is a theorem of ZF. Also, "$x$ is a finite ordinal" and "$x=\omega$" both are absolute properties according to the above Wikipedia reference. And, where does my proof break down? – Alex Eustis Jul 23 '24 at 13:29
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    My answer was a heuristic for general logic and axiomatic systems and models within. Jech concerns himself with transitive models in that particular statement. You can already see that cardinals are not absolute from his exercise 12.6 recommended after Lemma 12.10, where he claims in particular that $|Y| = |X|$ is not absolute for fixed $X$ (whether finite or not.) If you take a look at exercise 12.6 you'll see the "arrow" I referred to.

    In particular your induction proof breaks down at the base case where you assume $|Y| = 1$ is absolute.

     – TheAlertGerbil Jul 23 '24 at 15:07
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    @TheAlertGerbil Where does Jech say $"|Y|=|X|$ is not absolute for fixed $X$ (whether finite or not)?". I see he says $|Y|=|X|$ is not absolute in general, but you seem to be putting words into his mouth. – spaceisdarkgreen Jul 23 '24 at 15:20