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Some days ago I had the opportunity to listen to the talk about model theory and connections with algebra and geometry. I'm not at all specialist in this field so my question probably will be naive but nevertheless I'll try to explain my doubts. As far I understood: for example if we consider the theory of groups a model for such a theory will be a concrete group. For example the statement: "there is some $g \in G$ such that $g^2=e$" is a true statement in every model ($g=e$ is ok) therefore this statement is a theorem in a theory of groups. On the other hand the statement "for all $g \in G$ we have $g^2=e$" is no longer a theorem from the theory of groups: one can find examples (models) of the theory where this is satisfied but also one can construct counterexamples. In this sense this statement is independent from the axioms of group theory. So far everything looks clear. But I have a problem in understanding how the situation looks like in the context of set theory and ZFC axioms: for example I know that the statement "the cardinality of the reals is the next cardinality after the cardinality of the set of natural numbers" is independent from ZFC axioms. In other words one can construct two different models for set theory where in the one model this statement is true but in the second is false. What exactly does it mean "to construct model for set theory"? Let me return to the previous example about groups: for group theory a model is a concrete group so "to construct the model" means nothing more than "provide an example" but how to understand this in the context of whole set theory? What exactly do we have to construct? I'm sure that this question will be naive from the point of view of the specialist: from the other hand I suspect that there are at least few people which would like to know the answer for such "meta" question.

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    $\begingroup$ Dude. Paragraphs. $\endgroup$ – Asaf Karagila Dec 6 '15 at 5:03
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To construct a model of set theory means to produce a set $A$ and a relation $R$ on $A \times A$ such that all the axioms of ZFC are satisfied if we take "set" to mean "element of $A$" and take "$a \in b"$ to mean $aRb$.

This is not actually any different than the case with groups. The signature is different, and the axioms are different, but the definition of "model" is the same.

There is one complication, though. Although ZFC proves that there is a model of the group axioms, ZFC does not prove that there is a model of the ZFC axioms. One way we can get around this is by moving to a stronger system of set theory to construct the model of ZFC. For example, Kelley-Morse set theory proves that there is a model of ZFC. Another way is to simply assume there is one model, and use that to construct other models.

Sometimes, in set theory, we allow a more general kind of model in which $A$ is a proper class and $R$ is a definable relation on pairs of elements of $A$. These are called "class models".

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  • $\begingroup$ Thank you for your answer. So according to what you wrote: none model (understood in the first sense, not as "class model") could provide a theory which allows us to consider sets of arbitrary cardinality? I think that this is precisely why I didn't see an analogy: for me a set theory was always "the whole theory with all sets, where I could take arbitrary big sets"-such a theory requires the class instead of a single set as "the universe". $\endgroup$ – truebaran Dec 5 '15 at 23:06
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    $\begingroup$ "Inside" the model, there are sets of arbitrary cardinality. It will have a set it believes is $\mathbb{R}$, and the powerset of $\mathbb{R}$, etc. From the point of view of the model, the collection $A$ is a proper class. Of course, from outside the model, we see that the model only contains some sets, not all of them, and $A$ is a set. I think that one issue here is that we are used to looking at many different groups, so we don't have any qualms when we see different groups. But, outside of set theory, we don't normally look at different models of set theory, so it takes getting used to. $\endgroup$ – Carl Mummert Dec 5 '15 at 23:53
  • $\begingroup$ Ok, I think that I understood the idea. You are totally right with your last remark, it's hard to think from the outside of set theory. Many thanks! $\endgroup$ – truebaran Dec 6 '15 at 0:24

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