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Skolems Paradox shows an ostensible conflict between Cantor's Thoerem (CT) and the downward Löwenheim–Skolem Theorem (ST).

CT: for any set $A$, the powerset of $A$, $P(A)$, has a strictly greater cardinality than $A$. Cardinality is in terms of bijections: two sets have the same cardinality iff there exists a bijection between them. A set $A$ is countable iff there exists a objection between $A$ and the set of naturals $\omega$. Since (by CT) no function surjects $\omega$ onto its powerset, we learn that $P(\omega)$ is uncountable. Thus CT generally tells us some sets are uncountable.

ST: If a countable first-order theory has an infinite model, it has a countable model. The standard axiomatization of set theory, ZFC, is such a theory.

Assume ZFC has a model (which must be infinite).

By ST, ZFC has a countable model .

By CT, we can deduce the existence of uncountable sets from ZFC.

Therefore, there must be some set A ∈ such that satisfies $A$ is uncountable. That is, there is no bijection $f \in$ between $A$ and $\omega$.

However, since has a countable domain, there are only countably many elements available to be members of A.

Thus A appears both countable and uncountable.

Some good links I found:
1. How did first-order logic come to be the dominant formal logic? (and comments)

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pichael
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Indeed $\mathfrak M$ sees only a small fraction of the universe, in particular he does not see the bijection between $\omega$ and itself, or between $\omega$ and some of the members of $\mathfrak M$.

However it is perfectly fine to have a model which is countable and has elements which are uncountable.

If you start by taking the real numbers and using it to generate a countable model, you will end up with a countable model of ZFC (of course it would not know of its own countability) in which there is a really uncountable set.

Now recall that $\mathfrak M\models A\text{ is countable}$ if and only if there is $f\in\mathfrak M$ which is a bijection between $A$ and $\omega$. If $\mathfrak M$ does not know about such $f$, $\mathfrak M\models A\text{ is uncountable}$. This is the heart of internal-external points of view.

If, on the other hand, we require $\mathfrak M$ to be transitive then every element of $\mathfrak M$ has to be countable as well (since it is a subset of $\mathfrak M$ by transitivity).

Asaf Karagila
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  • I changed the OP a bit, is it OK? I fear "That is, A in is (at most) countable" might be wrong UNLESS I stipulated that was transitive. – pichael Jun 04 '12 at 18:45
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    @pichael: It is fine, however as long as you do not assume that $\frak M$ is transitive there is no guarantee that every element of $\frak M$ is countable. – Asaf Karagila Jun 04 '12 at 18:50
  • So basically you have been saying that my explanation assumes that $M$ is transitive. What would I have to add/subtract from my explanation to include non-transitive models of ZFC? – pichael Jun 04 '12 at 23:23
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    @pichael: If you look at the Skolem-Lowenheim theorem you will see that it does not only guarantee us a countable model, but also it allows us to point at a certain element and ensure that this element would be in the model. Now suppose we have a model in which there is a truly uncountable set $A$; we can consider the countable model in which $A$ is a member. Obviously this model cannot be transitive, as that would imply $A$ is also a subset of that model, which is a contradiction. So you have $A$ which both the universe and the countable model agree is uncountable... – Asaf Karagila Jun 05 '12 at 04:46
  • I thought we would have to distinguish between an $A^M$ and $A^V$, where according to $V$ $A^M$ is obviously (at most) countable, and where according to $M$ $A^M$ appears uncountable (as there is no relevant bijection). $A^M$ is a proper subset of $A^V$ that includes only the members of $A^V$ that M recognizes (since $M$ views itself as a proper class and can recognize no sets outside itself). That's what I have been thinking. – pichael Jun 05 '12 at 04:53
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    No, the set $A$ is the same set. The cardinality of the set changes. It may be the case that the set has some specific role (being $\omega$, being $\omega_1$, being $\cal P(\omega)$, etc.) which may also change between models. However as sets these are the same, especially when you take submodels of "good" models (i.e. transitive and well-founded models, which may be very large and not just countable). – Asaf Karagila Jun 05 '12 at 04:55
  • I just majorly edited the question. What I wrote is pretty much exhaustive of what I understand about Skolem's Paradox. The talk of transitive models, etc., I have only the slightest handle on, if any at all. Thanks for all the help. – pichael Jun 05 '12 at 05:55
  • Is the following correct? The "paradoxical" statement that there are uncountable sets in countable models of ZFC must be said from at least two perspectives. If $M$ is a countable model of ZFC, then as we have discussed, it can't say of itself that it is uncountable. However, $M$ is certainly saying that some set $A\in M$ is countable. Thus there are at least two perspectives. The reason why I add "at least" is because $M$ can be countable from any model $N$'s perspective that contains a bijection between $M$ and $\omega^N \in N$. – pichael Jun 13 '12 at 22:06
  • @pichael: I'm not sure what you are trying to say here. – Asaf Karagila Jun 13 '12 at 22:10
  • Skolem's Paradox is usually surprising, I suppose, because people initially assume it is being said from a single, absolute perspective. I'm trying to say that if $M$ is a model of ZFC, then the paradoxical statement—-there are uncountable sets in countable models of ZFC--must necessarily be said from two perspectives. Once someone understands these two perspectives, the paradox goes away. "$M$ is countable" is said from an external perspective on $M$, while "M satisfies $A$ is uncountable" is obviously said from $M$'s inside perspective. – pichael Jun 13 '12 at 22:13
  • @pichael: Yes, that way it makes sense. I suppose it's the late hour that's getting to me, but the first comment is very difficult to understand. The second comment, however, is crystal clear. – Asaf Karagila Jun 13 '12 at 22:15
  • I was just seeing if it was said from ONLY two perspectives or from AT LEAST two perspectives. I was going for the latter in the "unclear" statement because $M$ may be countable from many different model's perspectives just in case they have a bijection between M and their set of naturals. – pichael Jun 13 '12 at 22:17
  • @pichael: Yes, $M$ itself may be countable from the aspect of different models. – Asaf Karagila Jun 13 '12 at 22:18
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A set is countably infinite if it is in bijection to $\mathbb{N}$. If you have a countable model, a set can be uncountably from the "point of view" of the model simply because the model doesn't contain the bijection. So the model doesn't see the bijection, and a set is seen as uncountable.

Michael Greinecker
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  • Can I give a kind of example for your comment: Let M be a model that satisfies "A is countable." As you say, this just means there is a bijection in M consisting of N and A. Let's throw away all such bijections between A and N and call this new model M'. Thus M sees A as countable, M' sees A as uncountable. – pichael May 25 '12 at 10:49
  • @pichael: Of course you have to be careful when you say such things. If $A$ is a set which is provably countable (for example, $\omega$ itself) then $M'$ would not be a model of ZFC. – Asaf Karagila May 25 '12 at 10:50
  • @AsafKaragila: I must be missing something crucial here, as that is not immediately obvious to me. What is M' not satisfying? Pairing? – pichael May 25 '12 at 10:59
  • @pichael: $\omega$ itself is missing. It is definable (as an object) from ZFC therefore in every model we need to have something which acts as that set. Indeed it could be a different set. However if you require $M'$ to be a submodel of $M$ then $\omega$ must remain the same. – Asaf Karagila May 25 '12 at 11:01
  • @AsafKaragila: what if A was the rationals? or integers? – pichael May 25 '12 at 11:05
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    @pichael: You will note that I refrain from using $\mathbb N$. This is because $\mathbb N$ is a mathematical idea which is interpreted often as a model of PA which in ZF is interpreted canonically as $\omega$ along with ordinal operations. The same goes for $\mathbb Z$ or $\mathbb Q$ and so on. All of these can be interpreted as any countably infinite set. $\omega$, on the other hand, is a very concrete set in ZFC - it is the collection of finite ordinals. And yes, everything definable from $\omega$ also has to stay if you want $M'$ to be a submodel. – Asaf Karagila May 25 '12 at 11:08
  • @AsafKaragila: what does "everything definable from w" mean? My guess is that you're saying in order to be a model of ZFC, N, Z, Q, etc. all have to be in the model (as these are "definable from" w?). But, is the bigger picture comment here that I can't throw away the bijections as carelessly as I did while expecting M' to still be a submodel of M? – pichael May 25 '12 at 11:19
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    @pichael: It means sets which can be defined from $\omega$. For example the set of finite non-zero ordinals; the ordinal $\omega+1$; the set of ordinals which are decomposable in multiplication; etc. – Asaf Karagila May 25 '12 at 11:23
  • @pichael: Note that the natural numbers and the integers are both countable. You can define a binary operation $\dot+$ such that $(\mathbb N,\dot+)$ would be isomorphic to $(\mathbb Z,+)$. You may want to begin with understanding what does it mean for something to be definable. Perhaps start by reading this Wikipedia entry. – Asaf Karagila May 25 '12 at 11:53
  • @AsafKaragila: Thanks for the link and the help. I'll have to check it out tomorrow... – pichael May 25 '12 at 11:56
  • @AsafKaragila: Quick question though about the OP: is this an OK way to explain Skolem's Paradox? Anything wrong? missing? unnecessary? Again, many thanks. – pichael May 25 '12 at 11:58
  • @pichael: It would depend on the crowd. I suppose it is a good explanation if the crowd is made of people knowing a bit about logic - but not too much about logic. However in the year 2012 it cannot go unaccompanied by its solution. – Asaf Karagila May 25 '12 at 12:02
  • @Michael: I'm terribly sorry for hijacking this thread like this! :-) – Asaf Karagila May 25 '12 at 12:03
  • @Asaf: No problem. But I was slightly puzzled when I saw that I had 13 notifications in my inbox... :-) – Michael Greinecker May 25 '12 at 16:45
  • I'm sure you're probably exhausted from this thread, but maybe I could ask a beginner type question. How can a model itself be countable and yet not have a function which is a bijection with $\omega$? Thanks, –  Jul 08 '18 at 21:36
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    @Andrew The point is the meaning of a model "having a bijection with $\omega$" or plainly "what is a model". You might think of the model having a list of names for objects and not having a name for the bijection. – Michael Greinecker Jul 09 '18 at 16:12
  • Thanks, Michael –  Jul 10 '18 at 18:52