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Skolem's Paradox tells us that countability in first-order logic is relative.

Relative to what?

Below is what I've gathered.

Countability it relative to:
1. what a model takes to be $\mathbb N$
2. what bijections between $\mathbb N$ and some set $A$ a model recognizes.

Two examples:

For (1): Let $$ be a model such that $A$ is uncountable in $$. We add a bijection between $A$ and $\mathbb N$ to $$ and call this new model $$. $A$ is countable in $$. As mentioned here.

For (2): The underlying set of a model $$ might be countable from the perspective of a larger model $$, and so $$ might "see" $\mathbb N$ differently than $$. I'm not sure if I've said this right, but here is a post on this.

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pichael
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1 Answers1

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I fear I may have confused you a bit, but this is a confusing topic after all, and it can take quite some time to wrap your head around it completely.

First let us establish the following fact. We live in a big big universe. This universe, for the sake of conversation is a model of ZFC. However this universe is not a set, and we do not know any sets outside our universe.

This universe judges with (extreme prejudice) what is truly countable and what is not, what is countable is what the universe know has a bijection with $\omega$ (which in our case is "the true $\omega$").

Suppose that there is a model of ZFC $\mathfrak M$ which is a set in our universe, it may be countable and it might not be. It might know the same true $\omega$, and it might think that some other set is $\omega^\mathfrak M$ (the set which $\mathfrak M$ thinks is $\omega$). It is important to know, $\omega^\mathfrak M$ may not even be countable! In such case $\mathfrak M$ may think that things are countable even if they are not, as it compares things to its own $\omega$ (which we know is uncountable).

Since $\mathfrak M$ is small, it may know some sets which are truly countable, but it may not know about the bijections these sets have with the true $\omega$, it may be the case that $\omega^\mathfrak M$ is itself uncountable (but $\mathfrak M$ is unaware to this fact, since it judges countability wrong) and then $\mathfrak M$ will get "most" things wrong about countability.

So we end up with the following situation:

  1. There is an absolute notion of countability. This is what the universe decides, or knows, is countable.
  2. Every model inside the universe has its own version of $\omega$ which may be the true $\omega$, may be a different countable set, and in the worst possible case may not even be a countable set! Inside such model, $\mathfrak M$, a set $A$ is countable if the model knows about a bijection between the "local" $\omega^\mathfrak M$ and $A$.
  3. We can then extend such $\mathfrak M$ to a slightly larger $\mathfrak N$ in which some set $A$ which in $\mathfrak M$ was not countable, $\mathfrak N$ thinks is countable (we added the needed bijection).

We separate the case of 1, where the countability is absolute (or "true") even if internally some model $\mathfrak M$ may not know that some set is countable, from the cases of 2 and 3 in which a certain model thinks of a set as countable, or uncountable, regardless of its true size.

Asaf Karagila
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  • Technically, we can't say that there is such a universe (or a model of ZFC) because we don't know that ZFC is consistent, right? But assuming it is, by Completeness (?) there is a model/universe. And assuming there is such a universe, there would be an absolute notion of countability. – pichael May 29 '12 at 21:24
  • Second, so there can be a "truly" uncountable set A which our M takes to be countable?! But lets say that M is countable. Clearly, M doesn't then recognize the "true" A as there are only countably many members in M available to be in A (in M). (But then wouldn't A in M be a proper subset of the "true" A?) – pichael May 29 '12 at 21:32
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    @pichael: For the first comment, yes. We cannot know that ZFC is consistent, and we have to assume that and thus it has a model. For the second comment, $M$ can be a model generated in some way that ZFC is true in $M$ (i.e. $M$ is a model of ZFC) but the $\omega^M$ is uncountable. This means that things which $M$ think of as countable may be uncountable. Lastly, if $M$ is not transitive then its elements are not subsets of the model, that is we can have some $A\in M$ such that $A\nsubseteq M$. In particular we can have $M$ countable, $A\in M$ a truly uncountable set, thus $A\nsubseteq M$. – Asaf Karagila May 29 '12 at 22:03
  • So, the only way some A can serve as w in M while A is uncountable is that M is not transitive? But as we have talked about before (unless I misunderstood), M "sees" itself as a (proper?) class and thus doesn't recognize any sets outside M. So, to M, A is countable (obviously bc A is w in M) even though A is uncountable, right? What would happen if M was transitive? I've seen "transitive" come up in discussions of Skolem's Paradox, but I don't know about transitivity—e.g. here. I'll do some research. – pichael May 29 '12 at 23:52
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    @pichael: Transitive models are "nice" because they are closed under $\in$, namely $m\in\mathfrak M$ and $x\in m$ then $x\in\mathfrak M$. I'm not sure to answer you whether or not transitive models must agree with the universe on $\omega$, though. – Asaf Karagila May 30 '12 at 08:39
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    @Asaf: Suppose $\mathfrak M$ is transitive and doesn't agree with the universe about $\omega$, Then its $\omega$ would have to be larger than ours, because every element of our $\omega$ has a finite description in the formal language that proves it is in $\omega$. But since transitive models are well-founded, looking in from the outside we can see that there must be a least one among the non-standard elements of $\omega^{\mathfrak M}$. Such an element cannot have a predecessor, not even inside $\mathfrak M$. Therefore so it cannot be a member of $\omega^{\mathfrak M}$. Contradiction! – hmakholm left over Monica May 30 '12 at 13:59
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    On the other hand, if our universe is ill-founded enough for the Mostowski collapse lemma to be true for arbitrary models, we can adjoin a non-archimedean natural to ZFC (with or without Reg) by the usual compactness construction, get a model from the completeness theorem, and then collapse that in order to get a transitive (yet non-well-founded) model that disagrees with the universe on $\omega$. – hmakholm left over Monica May 30 '12 at 14:03
  • @Henning: That is great! – Asaf Karagila May 30 '12 at 14:20
  • @Henning: To your first comment. I don't get why w has to be larger than the universe's (I don't get the "because..." part.) I know "being well-founded" means there is a least element. Is a non-standard element, an element of w in M that doesn't agree with the universe's w? Then I'm lost with the remaining inferences. Is there a "for dummies" version? :) – pichael May 31 '12 at 00:48
  • @Asaf: "First let us establish the following fact. We live in a big big universe. This universe, for the sake of conversation is a model of ZFC. However this universe is not a set, and we do not know any sets outside our universe." This is just V, right? And V is a proper class? Proper classes can't be models, as models are sets. But we are just going to assume it is for the sake of understanding? OK. – pichael May 31 '12 at 01:53
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    @pichael: Classes can be models, these are called class models. We have to do set theory inside a universe of sets. Much like addition of integers lives "inside" $\mathbb Z$, despite the naive "all numbers are complex" being a larger universe - much like this is set theory. We take our sets from somewhere, and this somewhere is not a set for itself. We call this meta-theory, and it plays a role in mathematics (a role many people are oblivious to). My first sentence merely suggests that this meta-theory is ZFC, and therefore the universe is the inside of some model of ZFC. – Asaf Karagila May 31 '12 at 05:33
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    One important caveat about class models, since they are "too big" to be sets they do not adhere to the completeness theorem's requirement. Namely, if ZFC has a class model it need not imply that it is consistent. – Asaf Karagila May 31 '12 at 05:34
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    @pichael: You may also be interested in this answer. – Asaf Karagila May 31 '12 at 06:05
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    @pichael: Under the assumption the the model's $\omega$ is different from ours, it has to be different by being larger, because each of our natural numbers need to appear in the model's $\omega$ too. The "for-dummies" version is that if a model of ZFC is transitive, then its $\omega$ is identical to the $\omega$ in the outside universe. – hmakholm left over Monica May 31 '12 at 14:20
  • @Asaf: So model theory can be done just when there is an established meta-theory. Our meta-theory is ZFC, while it could be others (such as Z, ZF, MK, KP,...?). "and therefore the universe is the inside of some model of ZFC." Could this be stronger? E.g. "Our meta-theory (ZFC) determines what universe there is and thus what models are available to use." – pichael May 31 '12 at 19:33
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    @pichael: Mathematics is done when there is some meta-theory. Model theory, as it deals with sets usually requires us some well-defined notion of what is a set in the first place, so a very natural starting point is taking the meta-theory to be some sort of set theory. – Asaf Karagila May 31 '12 at 19:56
  • @Asaf: does M think that its w is w ^V or just w? (Or does w without V just make no sense?) – pichael Jun 01 '12 at 03:54
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    @pichael: Think of $\omega$ as a syntactical constant, the true $\omega$ is $\omega^V$, but $\mathfrak M$ does not know about that, it might know $\omega^V$ as its own $\omega$; and it might not know $\omega^V$ as such. There exists only one true [external] $\omega$, but every model has its own [internal] $\omega$ as well, some models may agree with the universe on what is $\omega$ and some models might disagree with the universe. That's all very confusing, and it takes quite some time to fully understand this. – Asaf Karagila Jun 01 '12 at 07:31
  • @Asaf: Exactly what in the OP indicated that I was confused? I.e. exactly what needed correction? Thanks. – pichael Jun 09 '12 at 05:21
  • @pichael: Wow, my memory does not have this sort of caching to answer you without reading everything. If you pinpoint me to a certain comment and be more specific in your question, I might be able to restore the wanted data... :-) – Asaf Karagila Jun 09 '12 at 07:59
  • @AsafKaragila: What?! All this time you had me believing you were omniscient... :P I was just referring to your actual answer (above), you wrote: "I fear I may have confused you a bit..." I was wondering what in my original question tipped you off about my being confused/what needed correction. – pichael Jun 09 '12 at 17:47
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    @pichael: Oh, that. Well, the fact that different models have completely different $\omega$ and that $\mathbb N$ is not $\omega$. One can easily restrict themselves to standard, transitive models of ZFC which agree on the real $\omega$. After understanding the paradox in those terms one can slowly wade into the stranger realm of the non-standard. In particular my insistence on separating $\mathbb N$ and $\omega$ can be confusing, I suppose. So when you wrote this question I felt that I may have confused you in my past answers. – Asaf Karagila Jun 09 '12 at 17:53
  • @Asaf: if $V$ is an absolute perspective, then why is countability relative? I realize this has to do with internal/external perspectives again. I guess I'm asking why truth (about some set) is relative to the inner perspective as opposed to $V$'s external perspective? – pichael Jun 10 '12 at 06:38
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    @pichael: Because we work inside the models of ZFC, not just in the universe. Inside the models we work as if we are unaware of the outside happening; however we - as omnipotent and omniscient beings of that model - are allowed to skip back and forth between the model and the true universe. From the inside it looks very large and from the outside it looks very small. – Asaf Karagila Jun 10 '12 at 06:51
  • @Asaf: How can models within $V$ have different versions of $\omega$? What's an example of how to models would come to differ on $\omega$? – pichael Jun 10 '12 at 18:12
  • @pichael: I was watching Blade Runner, but you started a new thread so I'll elaborate there. – Asaf Karagila Jun 10 '12 at 20:43
  • @Asaf: Good movie. No worries, I thought it was a bigger topic than just for a comment anyway. – pichael Jun 10 '12 at 20:44