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I just want to see if I am using the term "categoricity" correctly in the following context:

(1) I was thinking about why someone might reject a simple resolution to Skolem's Paradox.

(2) The resolution I had in mind was the following: model-theoretic truth is relative to what is in the model. Thus a set can be countable or uncountable -- it depends on the model. For example, it depends on whether a model is countable or uncountable.

(3) The rejection I had in mind: when we say that a model is countable, we are coming from a perspective of categoricity -- that is, a perspective from which we take countability to be absolute, or non-relative.

MY QUESTION: am I using "categoricity" right here?

pichael
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Let $\kappa$ be a cardinal. A theory $T$ is $\kappa$-categorical if any two models of $T$ of cardinality $\kappa$ are isomorphic.

A theory $T$ is categorical if any two models of $T$ are isomorphic. No (first-order) theory that has an infinite model is categorical.

As to the Skolem Paradox, the fact that a theory over a countable language, if it has an infinite model, has a countable model, is a nice result, with a quite accessible proof. A notable fact is that the result, for ZF, played a significant part in Cohen's proof that if ZF is consistent, then so is ZF plus the negation of the continuum hypothesis.

André Nicolas
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  • Categorical theory is often used to mean a theory that is $\kappa$-categorical for every infinite cardinal $\kappa$. – François G. Dorais May 27 '12 at 04:22
  • @AndréNicolas: "No (first-order) theory that has an infinite model is categorical." Right, someone that might have just been exposed to Skolem's Paradox, might not realize this immediately. They would probably have strong intuitions against it. I'm wondering if I can describe their intuitions as coming from a "categoricity perspective" -- one in which they assume that there is one model that determines whether or not some set is countable. I'm just wondering the term "categoricity" aptly describes their intuitive assumption. (This would also help me understand categoricity better.) – pichael May 27 '12 at 04:37
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    Some of the intuition about categoricity comes from second-order languages, which were, at least until well into the $20$-th century, the default background languages. – André Nicolas May 27 '12 at 05:05
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As Andre said, categoricity means that all models are isomorphic and $\kappa$-categoricity means that all models of size $\kappa$ are isomorphic.

The problem is that model theory is usually developed within a universe of ZFC. Things within think of that universe as a class, not as a set. If $\frak M$ is a countable model of ZFC living as a set inside a larger universe $V$, the model theory inside $\frak M$ does not have to coincide with the model theory inside $V$.

In fact this is true for class models, that is $V$ can be a "very large" model of ZFC, and there is a subclass of $V$ which is also a model of ZFC that is significantly "smaller" and these two models have different model theories (because they ultimately contain different sets).

When we say that a model of ZFC is countable, we come from an external point of view. That is we live in a very large universe and that universe happened to know a countable set $M$ with a binary relation $E$ such that $\mathfrak M=\langle M,E\rangle$ is a model of ZFC.

It is possible that $M$ has a subset $N$ such that $\mathfrak N=\langle N,E\cap(N\times N)\rangle$ is also a model of ZFC, and it is possible that these models are not isomorphic at all.

This tells you that ZFC is not $\aleph_0$-categorical, and that categoricity here is far from categoricity in philosophy.

Furthermore, Cohen's work on forcing shows us that countability is far from absolute. If $\frak M$ is a countable model of ZFC, then there is a "slightly larger" $\frak A$ which is a countable model of ZFC, but $\frak A$ thinks that $\omega_1^\frak M$ is countable. Namely we took a set that $\frak M$ thought is uncountable, and we added bijections between this set and $\omega$.

However, as I said before we do all these model theoretic considerations inside a fixed universe of ZFC (this universe is not a set in our context), and the question whether or not a set is countable is answered in this universe. The notion of countability, therefore, does not "go down" to models which live inside this universe. They have their own notion of countability.

Asaf Karagila
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  • "Furthermore, Cohen's work on forcing shows us that countability is far from absolute. If is a countable model of ZFC, then there is a "slightly larger" which is a countable model of ZFC, but thinks that ω1 is countable. Namely we took a set that thought is uncountable, and we added bijections between this set and ω." This is similar to what I wrote yesterday, but I instead removed bijections to show that a set can be un/countable relative to whether it is in the model. I can see why adding bijections is better than removing them. – pichael May 27 '12 at 06:57
  • @pichael: Yes, this is similar, but different because it may be impossible to remove bijections (if the model is minimal in some sense), but it is easier to add bijections. – Asaf Karagila May 27 '12 at 06:59
  • I get that countability is far from relative, and I get the internal-external point (you talked about also in other posts). I get that countability does not go down to submodels -- each model will have its own w as we discussed before. I was simply trying to make explicit the intuitions that one might initially have for (wrongly) rejecting countability's relativity. E.g., one assumes ZFC is alpha_null-categorical. Again, I know this intuition is wrong. However, if mathematical categoricity held for ZFC, then countability would not be a relative. notion in ZFC. Ya? – pichael May 27 '12 at 07:10
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    @pichael: If $M$ is a submodel of $N$ then they have the same $\omega$ (although not necessarily the same [internally] countable sets). What I said yesterday is that you cannot just pick any countable set and "throw away all the bijections". As far as initial intuition, I'm afraid that my memory is troublesome and I cannot recall how I originally felt about these things, I think I may have soaked them slowly along the way so when I was told about these things I was not as shocked as most people are. – Asaf Karagila May 27 '12 at 07:17
  • "If M is a submodel of N then they have the same ω (although not necessarily the same [internally] countable sets)." The conditional here just follows from M's being a model of ZFC right? It must satisfy, e.g. extensionality, empty-set, pairing, and infinity axioms. The parenthetical part of what you wrote is true because not all of the elements in N are in M, and therefore, this will affect the bijections with w. Is this right? – pichael May 27 '12 at 08:33
  • @pichael: Yes. Because they are rather closely related models of ZFC (one is a substructure of the other) they share (amongst other things) the same $\omega$, so it is the same set in both. Indeed the difference between them could be that we tossed out a few bijections and ended up with a smaller model. – Asaf Karagila May 27 '12 at 10:22
  • Hey! I thought tossing out bijections wasn't allowed! :P – pichael May 27 '12 at 10:53
  • @pichael: Yes, but if you built the larger model by adding a bijection to the smaller model... :-) – Asaf Karagila May 27 '12 at 10:59
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Usually, one assume that your theory of set theory is consistent and hence there is a model $V$ of set theory. If you take the formal theory of set theory to be ZFC, there exists a set $\omega^V \in V$. A set of $V$ is said to be countable if there is a bijection $f \in V$ between $\omega^V$ and your set.

Now you can do model theory within $V$. That is, the domain of your structure is an element of $V$. In particular, you can develop the model theory of ZFC inside of $V$. By Downward-Lowenheim Skolem, ZFC has a countable model in $V$. That is there is set $G$ which is countable relative to $V$ and an interpretation function (also in $V$) that is a model of $ZFC$. $G$ is countable in $V$ means there is a $f \in V$ such that there is a bijection between $\omega^V$ and $G$.

However, $G$ itself is a model of set ZFC. So there is something in $G$ called $\omega^G$. There is also thing called $\aleph_1^G$. Since ZFC proves that $\aleph_1$ is not countable, one has that $\aleph_1^G$ is not countable in $G$. That is there is no $f \in G$ such that $f$ serves as a bijection between $\aleph_1^G$ and $\omega^G$.

However in the overlying universe $V$ $G$ is countable and hence all its subsets are countable. However the the bisection that witnesses this countability is not an element of model $G$.

So when you are working with several models of set theory where one model may be an element of another, then it would probably be more clear to distinguish countability in $V$ and countability in $G$. Hence a cardinal $\kappa$ will always refer to the cardinal $\kappa$ of $V$.

As for categoricity, I am not sure what you want to know relative to the idea of the skolem paradox.

A theory is $\kappa$-categorical if there is only one model of the theory up to isomorphism. However, this is assuming you fix a universe $V$ of set theory that serves as the "place" where all models of your theory comes from.

William
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  • Thanks for the great exposition. Let me clarify my "perspective" talk: if ZFC was k-categorical, then countability would be "absolute." I.e., if ZFC was k-categorical, then there would be one "thing" we use to determine whether or not some set was countable. To my mind, V is not a proper "perspective," as V is not a set. -- I hope I'm not being hopelessly unclear, but I am just trying to describe why someone just shown Skolem's Theorem might reject it, namely, they assume ZFC is categorical, i.e. there is an "absolute perspective" from which we say whether or not something is (un)countable. – pichael May 27 '12 at 04:23
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    A model of ZFC need not have another model inside of it. It is consistent (relative to ZFC) that ZFC+$\neg$Con(ZFC) is consistent. That is a set model that contain no other set model. – Asaf Karagila May 27 '12 at 05:41
  • What is the difference between w^G and w^V? – pichael May 27 '12 at 11:15
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    ZFC can develop all the ordinals. So since $V$ is a model of $ZFC$, it contains sets $\alpha$ which the interpretation of the ordinals in $V$. $\omega^V$ is just what I denoted the set in $V$ that corresponds to the least infinite ordinals. $G$ is a different model of $ZFC$. Hence it has it own least infinite ordinal, which I denoted $\omega^G$. – William May 27 '12 at 11:30
  • @William: For some reason, I'm thinking (1) every model of ZFC will be a submodel of V and so (2) w is w and the superscript is superfluous. Clearly, I'm wrong, but I'm not totally sure why. Might it be because V "recognizes" different (or more) elements of w? I don't know... – pichael May 27 '12 at 11:46
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    @pichael: This is again an issue of internal and external. Note that the language has no constants or relations other than $\in$. However finite ordinals are definable and some infinite ordinals (like $\omega$) are definable. This means that their existence is provable from ZFC. It also means that every model of ZFC has an element which it thinks as $\omega$. If $V$ is a model of ZFC and $M$ is another model of ZFC which is also a set inside $V$ then $M$ need not know the "real" $\omega$, and need not be a submodel of $V$. – Asaf Karagila May 27 '12 at 22:06
  • @AsafKaragila: Ok, got it. What can we say about aleph_1^G in V? Also, is this answer all in the pretext of formulating ZFC in first-order logic? – pichael May 28 '12 at 00:18
  • It seems you are very interested in this topic. If you want to properly learn about Skolem's Paradox, I suggest you read the first chapter of Marker's $\textit{Model Theory} where he rigorously defines what a first order language and structure (model) is. Then take a peek into Jech for the model theory of set theory. – William May 28 '12 at 00:23
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    @pichael: I can't say anything. $\omega_1^G$ may be a countable set, it may not be a countable set, it might be what $V$ thinks is $\omega_1$ and it might be something completely different. – Asaf Karagila May 28 '12 at 06:18