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Looking into the open problem section of the book Model theory by Chang and Keisler, I noticed that many problems assumed semi-axioms like GCH. I talk about 'semi-axioms' because these "axioms" are shown (using informal ZFC) to be independent from the usual axioms ZFC.

How can one philosophically justifiy the usage of these axioms? Of course, it's legal to use any axiom one wants to use and one also can argue that in mathematics (being a "mind game", as I recall Asaf Karagila writing in one thread), one doesn't have to justify any axiom.

I have the following thoughts: Note that it depends one the philosophy of set theory one beliefs in. If one beliefs in truth platonism for arithmetic, then the usage of semi-axioms might be justified as follows: We use ZFC to conceptually to find a proof of the independence of a semi-axiom; then we convert this proof into a proof in Peano arithmetic, and since one beliefs in truth platonism of arithmetic, one can now be sure that ZFC + [the semi-axiom] is consistent. If one doesn't believe in determined truth values of arithmetical statements, then it's difficult to justify the use of these semi-axioms, since one then doesn't belief that statements such as "ZFC + [the semi-axiom]" (which is an essentially arithmetical statement) has a definite truth value! If one is a set-theoretical platonist, then the usage of semi-axioms is also problematic, since one then don't know, if one is using a true or a false assumption.

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closed as off-topic by Włodzimierz Holsztyński, Steven Landsburg, András Bátkai, Stefan Kohl, Marco Golla Oct 9 '16 at 11:22

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  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Włodzimierz Holsztyński, Steven Landsburg, Stefan Kohl
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    $\begingroup$ You don't need to believe in GCH in order to investigate a question of the form "does GCH imply...?". $\endgroup$ – Wojowu Oct 8 '16 at 20:30
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    $\begingroup$ The main use of GCH is that it resolves all of cardinal arithmetic, even in the singular cases. Since cardinality computations arise often in model theory, not always in interesting ways, assuming GCH is a good way to focus on the main subject and sweep away distractions. Another motivator is that GCH facilitates some transfinite constructions involving diagonalization. I'd be very surprised if the authors thought of the hypothesis as statement regarding foundational belief. $\endgroup$ – François G. Dorais Oct 8 '16 at 20:44
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    $\begingroup$ There are many papers in model theory where the main difficulty is in removing GCH-like assumptions from some arguments. As François indicates, the main use of the hypothesis is to simplify matters. One can later see what can be done without invoking it, and what results in a genuinely independent statement. $\endgroup$ – Andrés E. Caicedo Oct 8 '16 at 22:15
  • $\begingroup$ How can one philosophically justify the use of the commutativity axiom in abelian group theory? $\endgroup$ – Steven Landsburg Oct 9 '16 at 2:03
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It's worth pointing out that Chang & Keisler is a relatively old book (it was the first comprehensive textbook in model theory when it was first published in 1973). As Andrés points out in the comments, advances in model theory around the time the book was being written allowed set theoretic assumptions like GCH to be removed from many theorems. The most famous example of this is the Keisler-Shelah theorem, which states that two structures $M$ and $N$ are elementarily equivalent if and only if there is a set $I$ and an ultrafilter $U$ on $I$ such that the ultrapowers $M^I/U$ and $N^I/U$ are isomorphic. This was proven by Keisler in 1961 using GCH and reproven in ZFC by Shelah (using a much more complicated method) in 1971.

In modern "mainstream" model theory, is is extremely unusual to come across a paper in which GCH is used, and if you look through a more modern textbook like Marker or Tent & Ziegler, you will find very few set theoretic hypotheses, if any (though if I remember right, Marker includes a nice exercise or two about consequences of Martin's Axiom). This is partly because of the earlier effort in removing dependencies on set theory, and partly because, with the advent of stability theory, research interest turned away from explicitly set-theoretic questions about ultrapowers, two-cardinal theorems, and the like. On the other hand, set theoretic hypotheses are still quite common in adjacent areas, like the study of infinitary logics and AECs.

Finally, to address the question in the title: Model theorists are no more "free" to use "semi-axioms" than any other mathematicians. The situation is analogous to that in number theory, where many theorems are proven under the simplifying assumption of the Riemann hypothesis, but it is seen as a great improvement to provide a proof which does not rely on this assumption. If model theorists (at least those of the 60s and 70s) are more likely than other mathematicians to use set theoretic hypotheses, it is probably because they are more familiar with set theory and their field is closer to it.

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  • $\begingroup$ Excuse me, what are AECs? $\endgroup$ – Michael Greinecker Oct 9 '16 at 5:07
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    $\begingroup$ @Michael: Abstract Elementary Classes. $\endgroup$ – Asaf Karagila Oct 9 '16 at 7:00
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    $\begingroup$ Unfrotunately the organizers deciedd to close this question. I would like to add few remarks. ZFC is an incomplete theory, therefore there are matehmatically interesting statements that are not proveable from it. It is quite reasonable to consider mathematical statements that are consistent with ZFC, perhaps in the (near?) future after ZFC will be replaced or revised they might become theorems. $\endgroup$ – Rami Grossberg Oct 12 '16 at 13:40
  • $\begingroup$ @RamiGrossberg Indeed! By the way, it should have occurred to me that the questioner might be interested in these slides of John Baldwin's. $\endgroup$ – Alex Kruckman Oct 12 '16 at 14:35
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Even if you knew the "true" status of the continuum hypothesis, why wouldn't you be interested in results about the logical consequences of a set of axioms? I think that it is true that the hydra game always terminates. Even so, I find the fact that Peano arithmetic can't prove this is intensely interesting.

This is particularly true for GCH, since the way that its consistency is proven is very interesting. Gödel constructed within ZFC a certain natural model of ZFC, $L$, where the GCH holds. If we postulated that this model was in fact the true universe of sets, then that would settle the question. I have the impression that most set theorists who have an opinion think that $L$ is not the true universe, but that doesn't make them any less interested in the construction.

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  • $\begingroup$ Thanks! Assume that one is a platonist and thinks that every set-theoretic assertion is either true or false. Do I understand you correctly that you think that in this case, one can still "for fun" look at the consequences of ZFC + GCH? When I think about that idea, it makes sense, thanks! But isn't it then a little bit different from working with axioms all of which one is sure that they are true? $\endgroup$ – user99445 Oct 8 '16 at 22:27
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    $\begingroup$ They're still true, they're just relative to some model different from the one that you intended. This follows from the completeness theorem of first-order logic -- if you have a consistent theory, then that theory has a model. Whether or not that model is interesting is a matter of taste, not philosophy. $\endgroup$ – arsmath Oct 8 '16 at 22:36
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    $\begingroup$ There's also the fact that if GCH implies a model theoretic result, say, then it's a waste of time to look for a proof that it's false. $\endgroup$ – arsmath Oct 8 '16 at 22:40
  • $\begingroup$ @arsmath The thing I don't like about your last comment is that, in a Platonist setting, we may very well think that, say, $\mathsf{MA}+\lnot{CH}$ is true, and so we may very well be able to disprove model-theoretic consequences of $\mathsf{GCH}$. The point is that, for a Platonist, it is artificial to limit the assumptions of their theorems to $\mathsf{ZFC}$. $\endgroup$ – Andrés E. Caicedo Oct 10 '16 at 0:20
  • $\begingroup$ But it's still a waste of time to look for a proof in ZFC. If they think that CH is false, they have to actually suppose it. $\endgroup$ – arsmath Oct 10 '16 at 7:31

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