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I would like to understand the mechanism of following statement from this answer by Noah Schweber:

We could work in a non-$\omega$ model of ZFC. In such a model, there are sets the model thinks are finite, but which are actually infinite; so there's a distinction between "internally infinite" and "externally infinite."

How a set could be "internally", ie inside appropriate model be regarded as "finite", but be "externally" infinite? (btw, what means "externally"? An "overmodel" containing the submodel which regards this set as finite)

Eg, could somebody present an illuminative "toy example"? I know so far how construct models with "converse" phenomena, ie where eg the naturals $\omega^M$ of the fixed model $M$ (which tautologically trough the lens of this model are seen as countable) appear internally uncountable, see eg here for idea. Thank's to @Alex Kruckman for pointing out the correct phrasing of this phenomenon.
Here I'm also not 100 percently sure what is meant in by "externally" as contrast to internally (= inside fixed model) A guess: A kind of "distinguished overmodel" which contains all ordinals?

But the converse construction that some set is internally finite but externally infinite appears to me to be less plausible/ intuitive. Could somebody elaborate the phenomenon behind?

user267839
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  • A minor aside: "the naturals apear internally uncountable" is not a very accurate description of what's going on in the linked construction. There we have an object which externally is "related to" the naturals but internally isn't, so it's fairer to say "there is a set which appears uncountable but is secretly the natural numbers in disguise." Of course a model $M$ won't believe that (something it recognizes as) the naturals will be uncountable, by definition! – Noah Schweber Jul 23 '24 at 21:51
  • "Note, that I know that how construct models ... where eg the naturals apear internally uncountable". I suspect you have "internally" and "externaly" flipped here. Given a model $M$, the naturals of $M$, denoted $\omega^M$, are always internally countable by definition. What we can do is find models in which $\omega^M$ is internally countable but externally uncountable. – Alex Kruckman Jul 23 '24 at 21:51
  • @NoahSchweber: What do you mean precisely by "object which externally is "related to" the naturals but internally isn't"? (is implicitly a kind of "overmodel/true universe in background" fixed where we can relate certain object - in our case an ordinal - with the naturals in "usual way", via an "classical" inclusion of ordinals (-is this inclusion what you mean by "related externally"), which in turn in a spefic model might not exist? Is this the point you intend to emphasise above? – user267839 Jul 23 '24 at 22:53
  • @NoahSchweber: Also, could the phrase that certain fixed model "recognizes an object as blabla" - in example above a model "recognizing some set as naturals" - be made mathematically presise? What does mean this "recognizing" formally? – user267839 Jul 23 '24 at 23:03
  • @AlexKruckman: Thanks, that's exactly what I meant but poorly phrased. But - this clarifies to notion of "internal" - but what means in this context in turn "externally"? Like that such tautologically internally countable $\omega^M$ may be constructued to be "externally" uncoutable? Ie how "externally" contrasts here to "internally"? – user267839 Jul 24 '24 at 00:11
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    "Externally" refers to what's true, "internally" refers to what's true in the model. – Alex Kruckman Jul 24 '24 at 00:48
  • @AlexKruckman: "Externally" refers to what's true. Is true where? My guess: So far I understand this business with models of ZFC this plethora of models $M$ with "fake" natural numbers $\omega^M$ arise if we restrict ourself to first- order models of ZFC ( this is due to compactness theorem, right?) But in higher order models the model is unique, so it seems to be plausible to accept this higher order model of $ZFC$ as "universal/ distinguished", issentially this one we do think about math on undergrade level. Do you mean when you write that "something is externally true" by that it is – user267839 Jul 24 '24 at 10:11
  • true inside this fixed (...so at all choosen) but unique higher order model of ZFC? ( so it would make sense to talk in that setting about something like "externally" or "absolutely" (...even though I not like this word and tend to be careful with it) truth of statements. Is that what is meant by "external"? – user267839 Jul 24 '24 at 10:18
  • @user267839 There is no need to fix a model of ZFC to work in or to do any reasoning about higher order ZFC etc. You're welcome to do this, of course, but it's not necessary. Whenever we work in ZFC, we're reasoning using the axioms about a universe of sets. Whether it's a hypothetical arbitrary universe or the "true" universe is a matter of philosophy. In either case, the things I'm comfortable asserting are true (externally) are exactly the things I can prove! – Alex Kruckman Jul 24 '24 at 11:32
  • @AlexKruckman: So do I understand you correctly that the distinction in "$\omega^M$ is countable in $M$ (so internally)" vs that it is "uncountable externally" can be phrased as that it is "true inside $M$ that it is countable", vs. "it can be given a proof in ZFC that it is uncountable". So here "externally uncountable" = ZFC provable from its axioms that uncountable? Or do I misunderstand? – user267839 Jul 24 '24 at 12:21
  • @user267839 We can prove the following: $M$ is a model, $\omega^M$ is uncountable, and $M\models \text{"}\omega^M\text{ is countable"}$. – Alex Kruckman Jul 24 '24 at 12:26
  • @AlexKruckman: But isn't it strange? So far I understand it if a statement $S$ expressible inside language of a theory $T$ can be proved inside this theory, ie from its axioms, then it is true in every model of $T$, but in general the converse is wrong. But if we can proove $\omega^M$ is uncountable from ZFC axioms, wouldn't this property be inherited by all models of ZFC? What is the error in my reasoning? – user267839 Jul 24 '24 at 12:35
  • (So far I understand $M\models \text{"}\omega^M\text{ is countable"}$ means that the proposition "$\omega^M$ is countable" is true in $M$, ie the statement mapped to "true" wrt evaluation map of ZFC formulas to model $M$) – user267839 Jul 24 '24 at 12:44
  • More precisely what I mean and what confuses me is that say $S, S'$ are statements/ propositions expressed in language of a theory $T$ and $M$ is a model of $T$. $S$ is true in $M$ - notion $M \models S$ - means just that $S$ mapped to "true" wrt evaluation of functions, relations of $T$ in $M$. Note that this not means that $S$ is provable in $T$, just it's true in a model $M$. But in contrast if $S'$ can be proved from axioms $T$, then $S'$ is true in every model of $T$, right? Applied to above, this would – user267839 Jul 24 '24 at 12:55
  • imply $M\models \text{"}\omega^M\text{ is countable"}$ ( by choice of model) but also automatically $M\models \text{"}\omega^M\text{ is uncountable"}$ for even every model as well, due to what I wrote in prev comment, or not? Does it make sense? – user267839 Jul 24 '24 at 12:57
  • I'm glad you asked a new question - this comment thread was getting too long. I've posted an answer. Let's move the discussion there. – Alex Kruckman Jul 24 '24 at 14:22

3 Answers3

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Remember that "appear finite" just means "has an injection (in the model) into an element of the thing the model thinks is $\omega$." The point is that we can have a "nonstandard $\omega$," with elements which are - externally - infinite (but internally finite by virtue of literally being elements of the model's $\omega$). E.g. any model of ZFC+"ZFC is inconsistent" must be such a model, since it will need to have a "number" which codes a proof of a contradiction in ZFC and no such truly finite number exists (hopefully!).

I think this becomes much simpler if we shift from set theory to arithmetic. Consider a nonstandard model $\mathcal{M}$ of (say) first-order Peano arithmetic, $\mathsf{PA}$. We have the obvious initial segment embedding $i:\mathbb{N}\rightarrow\mathcal{M}$ whose image (by nonstandardness) is proper; any $m\in\mathcal{M}\setminus ran(i)$ is "externally infinite but internally finite." Set theory makes the picture messier by having lots of additional "stuff" going on, but it's ultimately the same.

Noah Schweber
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  • @RobArthan No, Alex was right - check the edit history. (Hence the ":P" in my previous comment.) – Noah Schweber Jul 23 '24 at 22:14
  • How can the formulation that a model $M$ "thinks" that certain object inside is is $\omega^M$, ie the "naturals of $M$" be formally phrased? – user267839 Jul 23 '24 at 23:33
  • Moreover, what do you mean by "a number, which codes a proof of a contradiction in ZFC"? A naïve guess: Is here some kind of philosophy involved which allows to express proof as "honest numbers"? – user267839 Jul 23 '24 at 23:38
  • You seem to have finally hit upon "truly finite" numbers and you may want to communicate this to Hamkins' oracle. – Mikhail Katz Jul 29 '24 at 06:57
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By Gödel's incompleteness theorem, there is a sentence $S$ of ZFC that encodes the assertion that ZFC is inconsistent and such that the theory $ZFC \cup \{S\}$ has a model $\cal M$. $\cal M$ must include a set $P$ that purports to be a proof of $S$ and so, as far as $\cal M$ is concerned, $P$ is a hereditarily finite set (because proofs are hereditarily finite structures), but $P$ can't actually be hereditarily finite because otherwise it would be a proof in ZFC of the inconsistency of ZFC. So tucked inside $P$, there is a set that is (externally) infinite, but that $\cal M$ (internally) thinks is finite.

Rob Arthan
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  • Could you clarify the part on " ...so far $\mathcal M$ is concerned, $P$ is hereditary finite set, but actually it cannot be (h.f. set). Could you elaborate a bit this beeing h.f. set concerning/ insidemodel $\mathcal M$ vs beeing actually not ("externally"?) – user267839 Jul 23 '24 at 23:53
  • As a non-standard model, $\cal M$ contains objects that purport to be positive integers but are greater than any of $1, 2, \ldots$. So $\cal M$ contains sets that it thinks are finite because they are in 1-1 correspondence with one of these non-standard positive integers, but in our metalanguage we know that such sets are infinite. – Rob Arthan Jul 24 '24 at 20:37
  • what I not understand, how can a 'bare' set $P$ in a fixed model of ZFC + ${S }$ be a "proof of $S$? A "proof" is a finite sequence of implications which are "legal" wrt the deduction system of the theory. A proposition in language of a theory is true in a model if it evaluates to "truth value" under $M$. But what do you mean by "that a set can be a proof"? Or did I misunderstood what you mean there? So I not understand this "set as proof" interpretation you seemingly invoke in your answer – user267839 Jul 25 '24 at 15:30
  • You can write down in the language of ZFC a formula $\phi(p, s)$ which in the standard model asserts that $p$ is a number representing a proof of a statement represented by the number $s$. (If this is unfamilar to you see https://en.wikipedia.org/wiki/Gödel_numbering.) In a model of ZFC together with the assertion $S$ that ZFC is inconsistent, there $\phi(p, s)$ holds for some numbers $p$ and $s$ such that $s$ represents $S$. The only way this can happen is for $p$ to a non-standard number. – Rob Arthan Jul 26 '24 at 21:47
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Here's a "toy model" with the feature you want, but phrased in the language of topoi. There are subtleties here about converting between these "sheaf models" and honest models of ZFC, but I'm going to ignore them to keep this answer simple${}^1$.

There's a topos called $\mathcal{M} = \mathsf{Set}^{\omega}$, and a "set" in this topos is actually the product of $\omega$-many "real world" sets (here I'm fixing some model of ZFC to consider "the real world"). Now one can compute $\mathbb{N}^\mathcal{M}$ is a product of $\omega$-many copies of $\mathbb{N}$, and $n^\mathcal{M} = \{ x \in \mathbb{N} \mid x < n \}^\mathcal{M}$ is the product of $\omega$-many copies of the real-world-set $n$. In particular, we living outside of $\mathcal{M}$ can see that $n^\mathcal{M}$ is uncountable, but it "looks finite" to people living inside $\mathcal{M}$. For any reasonable $\mathtt{isFinite}$ predicate you write, we'll have $\mathcal{M} \vDash \mathtt{isFinite}(n)$.

1: note that $\mathcal{M}$ is actually a boolean topos, so we don't need to worry about the usual subtleties here. You can think of $\mathcal{M}$ as a boolean-valued model taking truth values in the algebra of subsets of $\omega$. There are standard techniques for building an honest-to-goodness two-valued model from one of these, and truth in the two-valued model is closely related to truth in the boolean-valued model.

I hope this helps ^_^

Chris Grossack
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  • Just to clarify: The formulation that this $n^M$ "looks finite" to people inside $M$, what does this formally? A guess, ok you could say, this is equivalent to that there is an injection inside $M$ into $\Bbb N^M$ - the "naturals of $M$ - so it we regard latter as naturalls inside $M$ then it's plausible that its any broper subset "looks like" finite set. But then the question is why inside your model the $\omega$-product of $\Bbb N$ is exactly the object which should be regard as "the naturals" internal to $M$. This appears to be "plausible", but how to verify it formally? – user267839 Jul 24 '24 at 00:04
  • You can pick your favorite axiomatization of $\mathbb{N}$ (for instance, you might ask for a smallest model of PA, an initial algebra for $0$ and $\text{Succ}$, a smallest infinite set, etc). Whatever you choose, you can syntactically write down some theory $T$, and ask if $\mathcal{M} \vDash \ulcorner \mathbb{N} \vDash T \urcorner$. Similarly, you can write down any definition of finite you want (for instance, it injects into a member of $\mathbb{N}$, every injection to itself is also a surjection, etc) and we'll have $\mathcal{M}$ will think that definition is true of $n$. – Chris Grossack Jul 24 '24 at 02:23
  • As for how to "verify it formally", there's a completely mechanical translation you can do in order to turn questions about $\mathcal{M}$ into (more delicate) questions about "the real world". This is the business of forcing, and you can read more about how it works in the topos setting in, for instance, chapter VI of Mac Lane and Moerdijk's Sheaves in Geometry and Logic – Chris Grossack Jul 24 '24 at 02:25
  • you related the questions about some model say $M$ to those about "the real world": do I understand it correctly that here is implicitly the "first- order vs higher- order models" issue going on? Namely, the reason that we can produce a bunch of such models with "strange naturals" comes so far I know from compactness theorem available only for first- order structures. In turn there is no such result for higher order theories. So when you write "the real world", do you invoke - to keep it formally - a uniqueness result on higher order ZFC models to legitimize to declare one to the "real world"? – user267839 Jul 24 '24 at 10:34
  • This is essentially the "internal" vs " external" issue. Its clear to me that the notion "internal" is always relative to some fixed model, by "external" appears cumbersome, it appears that we give to some "model in background" some "universal meaning" in order to talk about "truth" in "real world". This leads me to what I wrote in prev comment, that seemingly some uniqueness issues must be involved to declare certain model to be "the real world" ( my guess is that this comes from higher order models, but I'm not sure about this) – user267839 Jul 24 '24 at 10:42
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    This example is interesting, but is a rather different phenomenon than the “non-standard”/“non-ω” models in the quotation OP is asking about. “Non-standard” (whether for models of first-order set theory, or arithmetic, or elementary topoi) is typically understood to mean that the numerals don’t exhaust the model’s naturals — categorically, that there’s some $1 \to N$ that’s larger than every numeral, or something similar to this. In this example (and any Grothendieck topos), $N$ is standard; what you’re observing here is that the global sections functor doesn’t preserve finiteness. – Peter LeFanu Lumsdaine Jul 24 '24 at 12:44
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    @PeterLeFanuLumsdaine -- I agree with what you're saying, but I wasn't trying to provide an example of a model with nonstandard naturals. I was answering this question "How a set could be "internally", ie inside appropriate model be regarded as "finite", but be "externally" infinite?", particularly the request for a "toy example" of this situation. I think it's easier to get your hands on a sheaf model like this and experiment than it is to do the same with a nonstandard model of PA. Plus, when I got here, Noah had already given a characteristically excellent answer through that lens! – Chris Grossack Jul 24 '24 at 17:16
  • @PeterLeFanuLumsdaine: Could you make precise what you exactly mean by a "numeral"? The point is that it seemingly includes a kind of circularity: Say we can naively those $m \in \omega^M $ (=" integers" of model $M$) as "numerals, for which there exists a $n$ with $S^{\circ n}$ where $S$ successor function. But in order to have welldefined composition $S^{\circ n}$ in this model $M$, we would need to be able to define before $S^{\circ n}$ for all $n \in \omega^M $, and thats circular, isn't it? Do you see how to resolve this problem with "numerals"? – user267839 Sep 03 '24 at 21:38