# Are there counter intuitive interpretations of ZF or ZFC?

Are there interpretations of ZF or ZFC that are non standard in the sense that $\epsilon$ is interpreted in a counter intuitive way that intuitively has nothing to do with "belongs to" or "is part of" (apart from formally modelling the axioms of ZFC)?

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Do you mean $\in$ instead of $\epsilon$? – Miguelgondu Aug 28 '14 at 22:19
@Miguelgondu Many books about set theory, particularly older ones, use $\epsilon$ to represent set-theoretic membership. For example, see pages 252 and following in General Topology of J.L. Kelley. – MJD Aug 28 '14 at 22:21
@MJD Didn't know that, thanks. They're both epsilon, am I right? – Miguelgondu Aug 28 '14 at 22:23
..I mena the only relation symbol in ZFC - name it $\rightarrow$, $\epsilon$, $\in$ or $blub$ – YonedaLemma Aug 28 '14 at 22:24
@Miguelgondu Yes, the $\in$ symbol is a stylized epsilon. – MJD Aug 28 '14 at 22:25

This could be along the lines of what you are looking for, although it is only a model for $ZF$ minus the axiom of infinity. Define a binary relation $\in_{\Bbb N}$ on the set of nonnegative integers $\Bbb N$ by $$\forall k,n\in\Bbb N\;\Big(k\in_{\Bbb N}n\Longleftrightarrow\text{the }k\text{-th binary digit of }n\text{ is nonzero}\Big)$$ Then $(\Bbb N,\in_{\Bbb N})$ satisfies all the axioms of $ZF$ except for the axiom of infinity.

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This is just the standard model of ZF minus the axiom of infinity (i.e., the finite sets), with its elements named in a nonstandard manner. – Dustan Levenstein Aug 29 '14 at 4:28
And the map is really just the witness that $\sf ZF-Inf+\lnot Inf$ is bi-interpretable with $\sf PA$. :-) – Asaf Karagila Aug 29 '14 at 10:01
Although the comments are correct, the answer is still relevant to the question. Notice that the question explicitly says what "non-standard" is intended to mean, and it's not "not isomorphic to a trnsitive standard model." – Andreas Blass Aug 29 '14 at 14:20

There is a notion of "standard model" for $\sf ZFC$. But for that we first need to understand two things:

1. Since set theory includes inherently infinite objects, there is no "natural interpretation" for set theory, like there is for Peano axioms, or like we have developed for $\Bbb R$ over all these years. There is a good reason to call these model "standard", and we'll get to that later.

2. Since we want to talk about models of set theory, we need to have some notion of set first, otherwise we can't really talk about models. So we will assume to be working within $\sf ZFC$. How can we work inside $\sf ZFC$ without having a model of $\sf ZFC$ first? We can either take it for granted that there is some large universe of sets that happens to obey the laws of $\sf ZFC$; or we can say that we are really writing down definitions and proofs, and we really work in some weak foundational theory, and what we are really proving is that $\sf ZFC$ proves that such and such is true and that we say that such and such definition is called a "standard model".

Yes. This is a big problem to swallow in one bite. If you are uncomfortable with this idea, then you might want to take a year or two to study set theory seriously, and study logic more seriously, and at some point this issue will resolve itself (as do most problems in mathematics: they resolve themselves after sufficient study).

So we are working inside a universe of sets, which happened to satisfy $\sf ZFC$, or some related theory. This means that we already have a notion of $\in$. The one that comes with that universe of sets. Let us call this universe $V$.

So a model of $\sf ZFC$ is a set $M$, in $V$, and some binary relation $E$ on $M$, such that $(M,E)$ satisfy all the axioms of $\sf ZFC$.

Now we can ask, if there is a set $M$ such that $(M,\in)$ (in which case I really mean $\in\restriction(M\times M)$) is a model of $\sf ZFC$. It might be true that such set exists, or it might be false. This will depend on $V$. As it was mentioned by others, $V$ might be such that there aren't any models of $\sf ZFC$. Since if we could prove that there is always such model, then $\sf ZFC$ would prove its own consistency, which we know it can't.

If such $M$ exists, that $(M,\in)$ is a model of $\sf ZFC$ then by Mostowski's collapse lemma, we can "collapse" it, and get a transitive set, namely every element of $M$ will be a subset of $M$. And this means that $M$ really perceives the notion of $\in$ as the universe does. And in that case we say that $M$ is a standard model of $\sf ZFC$.

And since there are standard models, there are non-standard models too. Any model which is not isomorphic to a standard model is called non-standard. The strange thing about non-standard models is that they are never well-founded (otherwise Mostowski's collapse lemma would allow us to make them isomorphic to a standard model). This means that if $(M,E)$ is a non-standard model, then there is a sequence $X=\{x_n\mid n\in\Bbb N\}$ of elements in $M$, such that $x_{n+1}\mathrel E x_n$. How does that sit well with the fact that $M$ satisfies the axiom of foundation? Well, $M$ is aware of the existence of each of the $x_n$'s separately, but it does not know the entire collection of the $x_n$'s. Namely, there is no set in $M$ which represents $X$, or in other words, there is no $Y\in M$ such that $X=\{y\in M\mid y\mathrel E Y\}$.

Non-standard models can be easily obtained from any model by taking ultrapowers or by using compactness. On the other hand, it is possible that there are no standard models, only non-standard models, much like it is possible that there are no models at all.

So what would be a counter-intuitive interpretation of $\sf ZFC$? I'd say that any non-standard model would be. As a result, this means that any interpretation which doesn't know of models of $\sf ZFC$ is also non-standard (since the statement "There is a model of $\sf ZFC$" can be translated to a statement about Gödel numbers, which is absolute between standard models, so if a standard model exists, it must know about models of $\sf ZFC$ itself).

There is a lot more to say about these things. There are delicate points, and common pitfalls that one would implicitly assume are true, but don't have to be. And there are ways to get around those as well. The standard interpretations, if so, would be those which don't have these pitfalls (or at least not the most of them).

One important addition that I'd still like to make, is that given any model of $\sf ZFC$, we can work internally to that model, and now that model is our universe, and we can do anything that $\sf ZFC$ proves possible within that model. So philosophically, you can view mathematics as something in a fixed universe of set theory, or a fixed universe of mathematics, or you can view set theory as something which has many different universes, which are different models of set theory in larger universes (which may, themselves, be models of even larger universes and so on).

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@MJD: And I pride myself for being good with counting parenthesis! For shame! :-) – Asaf Karagila Aug 29 '14 at 14:10
I think it is not accurate to say that $V$ may not be aware of the existence of models. Perhaps what you mean is that maybe there are no models. (If there are, they are in $V$. "Standard" is defined according to $V$, so there are no issues here that $M$ is a model but $V$ thinks it's not.) Mentioning "awareness" suggests provability, which is not an issue of what $V$ sees or not. – Andrés E. Caicedo Aug 29 '14 at 14:28
@Andres: By "aware" I mean, of course, that there is no set $M$ with a binary relation $E$, such that $V$ thinks that $(M,E)$ satisfies [what $V$ perceives as] $\sf ZFC$. Regardless to $E$ being $\in$ or not. – Asaf Karagila Aug 29 '14 at 14:33
Another minor thing: It is not $X$ that matters whether it belongs to $M$ or not (Since $x_n\in X$, but what matters is whether $x_n\mathrel {E} X$.) – Andrés E. Caicedo Aug 29 '14 at 14:36
@Andres: I've edited a bit the points you mentioned. Thanks. – Asaf Karagila Aug 29 '14 at 18:49

"is interpreted in a counter intuitive way that intuitively has nothing to do with "belongs to" or "is part of""

The main issue I have with what you wrote is that you are assuming there you know that the intuitive meaning of $\in$ is. What you wrote seem to suggest that you already know what "belongs to" or "is part of" is really suppose to mean? You are assuming whenever you do mathematics you are living in some particular universe of set theory. As we can not prove there are any models of set theory, a good question is whether there is any natural meaning for "belongs to".

The confusion is when exactly did you define $\in$. Consider two common symbols, $\emptyset$ and $\{\emptyset\}$. In common mathematics practice, the brackets seems to suggest that $\emptyset \in \{\emptyset\}$. However, why should this be? If we have not yet defined $\in$, $\emptyset$ and $\{\emptyset\}$ are a priori just two symbols. We defined that $\emptyset \in \{\emptyset\}$ either formally or by defining the convention of using brackets. Again similarly for why should $3 \in \mathbb{N}$ be true. Before working in actual model of set theory, depending on your definitions, $3 \in \mathbb{N}$ but you do not actually know what $3$ or $\mathbb{N}$ actually are so how could conclude it is natural that $3$ belongs to $\mathbb{N}$.

I am not sure if the example were clear at all, but I was trying to convey that belongs to is not an absolute concept. Your hand may naturally belong to your body, but for abstract objects in mathematics $3$ belongs to $\mathbb{N}$ is not natural but a consequence of how exactly you constructed $3$ and $\mathbb{N}$.

However, relative to a fixed models of set theory, the idea of a nonstandard definition of $\in$ does make sense. Let $(M, \in)$ be a fixed model of set theory. Any (proper class) binary relation $E$ on a proper subclass $N$ of $M$ will define another $\in$-structure denoted $(N,E)$. If $E \neq \in \upharpoonright N$, then (important) from the point view of $(M,\in)$, $(N, E)$ has a nonstandard $\in$-relation. These occasionally do come up in set theory in permutation models and ultrapowers of the universe by ultrafilters. It should be noted that if $(N,E)$ is such that $E$ is well-founded, then $(N,E)$ is isomorphic to a $(N', \in)$ where $\in$ is the same as it was in $M$. However $E$ does not need to well-founded, for instance this happens when you take ultrapowers by ultrafilters that are not $\sigma$-complete.

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One needs to interpret your question with care. Here is one way that gives a particularly unimaginative answer to your question. Suppose that $M$ is a model of set theory (any axiomatization you like). Suppose further that in it $\in$ has its usual meaning (however you want to interpret that).

Now, take any other set $X$ with a bijection $M\to X$. You can now use that bijection to reinterpret the axioms of set theory and obtain that $X$ is also a model. Of course, $X$ will not look anything like $M$ and $\in$ will not look anything like actual membership. It will only behave like one.

In short, using the very simple trick of changing the names of the elements in a set, any model of set theory can be turned into one where membership is really really weird when you look at the elements, yet it behaves perfectly ok. I hope this is inline with what you are asking.

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