9

I apologize in advanced if I'm hopelessly confused...

Skolem's Paradox, I suppose, can be put like this:

$M$ is a countable model of ZFC and $M$ implies the existence of uncountable sets.

I suppose that people find this initially paradoxical because they assume the statement is said from a single, absolute perspective. However, there are (necessarily) two perspectives involved: the inside perspective on $M$ and the outside perspective on $M$. Once these perspectives are separated, we realize that there is no paradox. Consider:

The former conjunct "$M$ is a countable model of ZFC" is necessarily said from an outside perspective on $M$ -- as discussed here. Actually, $M$ can't express its own cardinality at all.

(Is $M$'s inability to express its own cardinality related to $M$'s being a proper class -- namely, that there is no function in $M$ that takes one of $M$'s members onto the universe of $M$?)

Continuing on…Let $N$ be the outside perspective on $M$ such that there is a bijection $f\in N$ between the domain M of $M$ and $\omega^N\in N$.

The latter conjunct "$M$ implies the existence of uncountable sets" is obviously said from $M$'s inside perspective -- after all, $M$ is a model of ZFC and thus must satisfy Cantor's Theorem.

So, we can separate the perspectives of the paradoxical statement above here:

  • From $M$'s perspective, $M$ is a proper class and there is some $A\in M$ such that $A$ is uncountable in $M$.
  • From $N$'s perspective, $M$ is countable.

These two statements are jointly consistent when we realize that what can be said of a set $B$ is relative to what some model has to say about $B$. And so, the paradoxical statement isn't so paradoxical.

Is there anything wrong with what I have written above? (It took me a long time to learn this stuff, esp. with zero background in set-theory, higher-math, higher-logics, model-theory, etc., so specifically telling me where I am going wrong, if I am, will be a great help and a great relief.)

What I'm really interested in is what can we say about $A$ from $N$'s perspective? Are the following possible:

  1. $M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as finite.
  2. $M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as countable.
  3. $M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as uncountable.

Under what conditions might (1) - (3) be individually possible (obviously they can't be jointly possible)?

I suspect this question might be rather simple. For example, (2) could be possible when $N$ recognizes a bijection both between $M$ and $\omega^N$ and between $A$ and $\omega^N$. (3) could be possible when $N$ recognizes a bijection between $M$ and $\omega^N$ but doesn't recognize a bijection between $A$ and $\omega^N$.

My goal here is to understand/stress the fact that truth in model theory is relative to what particular models have to say about their members. So, I'm trying to see that while $M$ may take $A$ to be uncountable, $N$ can take $A$ to be of any cardinality even under the condition that $N$ sees $M$ as countable.

Community
  • 1
pichael
  • 899
  • 6
  • 13
  • Some time ago I found a senior thesis that I thought came from somebody at Arizona that discussed this well. Unfortunately, my search didn't find it. – Ross Millikan Jun 14 '12 at 03:57

2 Answers2

6

The second and third are easily describable:

Suppose that $N$ is a model of ZFC and $N$ thinks that $M$ is a countable transitive model of ZFC ($M$ may not be such model, but internally to $N$ this assertion holds).

This means that $N$ thinks that $M$ is countable, and that every element of $M$ is countable. $M$, on the other hand, knows some sets which are uncountable to it. So we have $\omega_1^M$ is a countable ordinal in $N$, so the second situation holds.

Suppose now that we have a nice model of ZFC, $N$, which is uncountable and it knows about uncountable sets. If we take $\omega_1^N$ we can consider $M$ a countable elementary submodel of $N$ such that $\omega_1^N\in M$. By elementarity $M$ and $N$ agree on $\omega$ and both agree that there is no bijection between that set and $\omega_1^N$. So we have the third situation in which both models agree that some set is uncountable.

Lastly to address the first situation, what we want is to have an ill-founded model of ZFC which thinks of an uncountable set as its $\omega$, but that it will also know about some model which is nicer. I am not sure how to address this situation since for $N$ to think that $M$ is a model of ZFC, $N$ would have to assert that $M$ have certain properties in $N$. These properties may make it impossible to make the jump from infinite to finite in this manner.

There is a possibility that $N$ will not be aware that $M$ is a model of ZFC, but that defeats the purpose because then we talk from an external point of view about both these models.

Asaf Karagila
  • 393,674
  • As you can see, I've poured most of our discussions in here. How's the first part? – pichael Jun 14 '12 at 07:48
  • @pichael: It looks reasonable, especially for someone lacking formal training in set theory. – Asaf Karagila Jun 14 '12 at 07:51
  • When and why would someone assume transitive models? Or if that is too broad a question, are transitive models typically assumed (everyone seems to make that assumption)? In my circumstance, writing a very brief paper (sketch, rather) on Skolem's Paradox (I've just gotten carried away on this site), is there any reason to/not to assume/stipulate transitive models? – pichael Jun 14 '12 at 07:55
  • @pichael: Transitive models are simply nicer, they share the same ordinals with the universe and this makes them behave very good. In forcing, for example, it is often common to assume we work with transitive models because it helps. However in general there is no reason a model is transitive, especially when looking for counterexamples like here. – Asaf Karagila Jun 14 '12 at 08:12
  • Link to something on countable elementary submodels. – pichael Jun 14 '12 at 08:54
  • Can $M$ satisfy "$M$ is a proper class"? Is this just saying that there is no function in $M$ that maps any element of $M$ onto $M$? – pichael Jun 14 '12 at 10:15
  • 1
    @pichael: This is quite easy to formulate in dozen of different ways, the first one coming to mind is similar to what you suggest, "for every $x$ and every function whose domain is $x$ there is $y$ such that $y$ is not in the range of $f$". There are other ways too. It means that "Can $M$ satisfy ..." is not the right choice of words: $M$ has to satisfy that. – Asaf Karagila Jun 14 '12 at 10:24
2

(CAVEAT LECTOR: this post was made with the implicit assumption that $M$ and $N$ are transitive models of ZFC; that makes several of the thoughts here inaccurate. See the comments below.)

The only possible situation here is (2). (1) is impossible because finiteness is absolute - a set that's finite is 'always' finite. (Whoops - see below!) (3) is also impossible: since $A\in M$, and since $N$ 'sees' $M$ as a countable transitive set, $N$ also has $A\subset M$; applying the bijection between $M$ and $\omega$ to $A$ then gives a bijection in $N$ from $A$ to a countable set.

As for the correctness of what you said, I think it's worth putting more emphasis on what countability and uncountability mean in models; I would speak more explicitly about the lack of any bijection $g$ from $A$ to $\omega$ in $M$ and how $N$ can have such a bijection that $M$ is 'unaware' of.

EDIT: as pichael points out in the comments, finiteness is not absolute; being a specific finite number is, and I believe being a finite ordinal ought to be, but just the claim 'A is finite' isn't absolute across models. On the other hand, for the specific case of $A = \left(2^\omega\right)^M$, it's still possible to show that $N$ can't think it's finite, because $M$'s inclusion of $\omega\subset A$ carries through into $N$, and $\omega$ is absolute.

Steven Stadnicki
  • 51,746
  • 1
    Thanks for answering. With regard to finitude being absolute, this post, if I am understanding it right, says otherwise. – pichael Jun 14 '12 at 06:34
  • The first paragraph is plain wrong. The first problem was pointed by pichael, the second problem is that you assume that $M$ is transitive (with respect to $N$). It is possible that $M$ has sets which $N$ does not know about; sets that $N$ does know about and are not subsets of $M$, that is $A\in N$ and $A\in M$ but $A\nsubseteq M$. – Asaf Karagila Jun 14 '12 at 06:59
  • @AsafKaragila I see now that the OP doesn't mention transitivity of $M$ (which I was implicitly assuming for no particularly good reason). Am I at least correct in recalling that transitivity itself is absolute? (i.e., if $M$ is a transitive model, then $N$ should still 'know' that $M$ is a transitive model; in other words, 'with respect to $N$' is redundant there) – Steven Stadnicki Jun 14 '12 at 07:16
  • 1
    @Steven: Suppose that $M$ is a countable transitive model (c.t.m), and $M'\in M$ such that $M\models M'\text{ c.t.m}$, now take an ultrapower of $M$ by some ultrafilter which is not countably closed. This would result in a model $N$ which is ill-founded and surely not transitive, but $N$ would know about $N'$ which is c.t.m with respect to $N$ (simply by Los' theorem). – Asaf Karagila Jun 14 '12 at 07:18
  • @AsafKaragila Ahhh, okay. I obviously need to bone up on my model theory at some point - I thought transitivity was absolute, but of course it's only absolute with respect to transitive models, and a non-transitive model doesn't have to agree on what is and isn't transitive. Requiring $M$ and $N$ both transitive I think salvages my points, but the argument is definitely strained. – Steven Stadnicki Jun 14 '12 at 07:31