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I hear that the axiom of choice (AC) derives from The generalized continuum hypothesis(GCH). And also hear that both AC and GCH are independent of Zermelo–Fraenkel set theory(ZF).

So, I'm just curious why don't expert mathematicians use ZF+GCH instead of ZF+AC(ZFC).

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    $\begingroup$ @Todd: Why is this question community wiki? $\endgroup$
    – Asaf Karagila
    Commented Oct 31, 2014 at 20:04
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    $\begingroup$ @Nate: For one, there is no measure on all of $\mathcal P(\Bbb R)$ which extends the Lebesgue measure (since the existence of such measure would imply that the continuum is weakly inaccessible, and in fact much much more). $\endgroup$
    – Asaf Karagila
    Commented Oct 31, 2014 at 20:06
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    $\begingroup$ @NateEldredge: I don't know whether you would consider this as a "working mathematician" statement but I think people might be using this "true" fact unconsciously all the time: Continuum function ($\kappa \mapsto 2^{\kappa}$) is strictly increasing. This is true under GCH but false under MA+~CH. $\endgroup$
    – Burak
    Commented Oct 31, 2014 at 20:14
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    $\begingroup$ @AsafKaragila It was suggested to make it so, and it seems to me that makes sense since there could be many reasons why mathematicians choose not to use ZF+GCH in place of ZFC, without there being one definitive reason. But I am prepared to hear counterarguments. $\endgroup$ Commented Oct 31, 2014 at 20:16
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    $\begingroup$ @Todd: It doesn't "feel" like a CW type question. But I don't have a better argument than that at the moment. $\endgroup$
    – Asaf Karagila
    Commented Oct 31, 2014 at 20:18

3 Answers 3

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In some sense $\sf GCH$ is a limiting axiom. While it solves a lot of things, it also means that certain things we are interested in become false or trivialized. And that's no fun.

For example, forcing axioms like $\sf MA$ become trivial assuming even just $\sf CH$, and stronger forcing axioms like $\sf PFA,MM$ and others become false (since they imply $2^{\aleph_0}=\aleph_2$).

So while $\sf GCH$ gives us more power in terms of implications, it also gives us more constraints. On the other hand the axiom of choice gives us a lot of power, since it helps tame infinite objects, but it leaves enough space to other axioms to develop.

And since you're asking, we can take this question one step further. $\sf GCH$ can be derived from the axiom $V=L$ (Godel's axiom of constructibility), so why aren't we just assuming it and that's that? Well, essentially the same answer. While it has a lot of merits, it also imposes a lot of constraints, more than we might want to assume.


Here's another (perhaps a better) argument in favor of $\sf AC$ and against $V=L$ and $\sf GCH$.

When working over $\sf ZF$ and you use forcing, then once you force $\sf AC$ to be true, it will remain true in every other generic extension, whereas $V=L$ is immediately false once you add any new sets to the model, and will never again be true if you only allow yourself generic extensions; and $\sf GCH$ is like a switch that we can turn on and off using [possibly class-] generic extensions.

So this is a very good argument in favor of the axiom of choice. It is stable under forcing, which is a lovely technique. On the other hand, limiting yourself to $\sf GCH$ makes forcing more difficult (because you will always have to ensure that the forcing does not violate that), and insisting on $V=L$ will make forcing outright impossible.

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Warning: naive answer follows. I think AC is more intuitively true than GCH. Also, I believe, AC is more indispensable for mathematics than GCH. For example, there are many equivalent forms of AC within ZF that come up naturally, such as the Zorn lemma or the well-ordering principle.

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    $\begingroup$ Yes, it's true. Also much more usable equivalents of the axiom of choice: Every vector space has a basis, every unital commutative ring has a maximal ideal, every set can be endowed with a group structure, etc. etc. $\endgroup$
    – Asaf Karagila
    Commented Oct 31, 2014 at 14:54
  • $\begingroup$ @AsafKaragila: Your response expresses better and more completely what I wanted to say. On the other hand, I don't agree with your comment above. While I certainly cherish that every vector space has a basis, I think Zorn's lemma is more useful as a mathematical statement (e.g. it implies easily that every vector space has a basis). Same for the well-ordering principle which allows us to assign a cardinality to every set, which I think is fundamental. $\endgroup$
    – GH from MO
    Commented Oct 31, 2014 at 18:22
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There are two distinct questions that you might be asking.

  1. Why has the mathematical community adopted ZFC as a standard foundation and not ZF+GCH?

  2. What mathematical and philosophical arguments can be advanced for and against adopting AC and/or GCH as a fundamental axiom?

Much as we might like to believe that the reason for the sociological acceptance of ZFC is that there are strong rational arguments for it, and that mathematicians accept those strong rational arguments because mathematicians are strongly rational, I do not believe that this is true. I believe that ZFC has been adopted as the standard largely for historical reasons. At some point, the mathematical community recognized the value of having some standard, because it would demonstrate that all of mathematics could in principle be formalized in a single system that avoided all the known paradoxes, and ZFC just happened to show up at the right place at the right time. I suspect that various other candidates could have "landed the job," including ZF + V=L, or even Z + C, and that ZFC was just a bit lucky.

After a choice was made, most of the mathematical community lost interest in foundations and so had no real interest in tinkering with this or that axiom to get "better" foundations. I don't think that ZF+GCH was ever a serious contender, because GCH was still considered an open problem by the time ZFC had already secured its status as "the" foundation. If a statement X is considered an "open problem" then people generally do not also consider X to be a candidate for a basic axiom. By the time the independence of GCH was established, it was too late to apply for the job.

Having said all that, I should add that if by "expert mathematicians" you mean experts in set theory, then the question is a bit different, because set theorists are more interested in these sorts of questions than the mathematical community as a whole is. Some of them will tell you that they reject GCH as an axiom simply because they don't believe that GCH is true. Still, even among set theorists, a sizable proportion take what you might call a "pragmatic" approach. They care mostly about whether the standard base theory is a technically convenient one for the investigations that they are interested in. Then the considerations that Asaf Karagila mentions come into play. GCH just isn't the most natural or convenient axiom to use for most things that set theorists currently care about. If it does happen to be useful in a certain context then they won't hesitate to assume it, but such occasions don't come up that often. (By the way, often Martin's axiom turns out to be what you really need when you might think you need CH.)

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    $\begingroup$ I'm not 100% sure that I agree with that "ZFC just happened to be there" argument. If I recall correctly Bourbaki used something more akin to ZC; but there were critics of Zermelo's original axioms within the set theory community, and it took a while before ZF(C) was proposed as a whole unit, not to mention that by that time set theories like NBG were beginning to emerge as well, which could have been used instead of ZFC just as well. No, this was more than just lucky timing. $\endgroup$
    – Asaf Karagila
    Commented Nov 1, 2014 at 7:43
  • $\begingroup$ I don't like the "ZFC just happened to be there" argument. The close relationship between ZFC and NBG, together with some ideas behind NBG like "global choice", "limitation of size", and "single sorted vs. second order" single out ZFC as a really nice compromise. The candidate "ZF + V=L" is an attractive set theory, but if you are allowed only "one" set theory, then this set theory is too limiting. The wish to have only "one" set theory also makes "Z + C" or "bounded Z + C" unattractive, even if they are not as explicitly limiting as "V=L". $\endgroup$ Commented Nov 1, 2014 at 9:17
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    $\begingroup$ Asaf and Thomas: Well, this is a matter of opinion. However, I think that you underestimate the sociological factors. The mathematical community needed a foundation, ZFC was the best choice at the time, and then the case was essentially closed for reasons that were not purely technical. Non-set-theorists lost interest. If they hadn't, and the debate were still alive today (which it is actually threatening to be, due to recent developments in proof assistants, not to mention homotopy type theory), a different foundation might have supplanted ZFC. $\endgroup$ Commented Nov 2, 2014 at 19:38
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    $\begingroup$ Conversely, non-set-theorists could have lost interest earlier and settled on something less satisfactory than ZFC, leaving the set theorists complaining but essentially unable to do anything about it, just like many other unfortunate but entrenched mathematical conventions. $\endgroup$ Commented Nov 2, 2014 at 19:40
  • $\begingroup$ I'm not sure ZFC came along at just the right time. Russell-Whitehead-style type theory was already a fairly standard foundation, and was expected to serve as a foundation for all of mathematics (although only part of that claim had been verified) for quite a while after ZFC was available but before ZFC became the generally accepted foundation. Note, for example, that the title of Gödel's paper on incompleteness explicitly mentions "Principia Mathematica und verwandte Systeme". $\endgroup$ Commented Nov 15, 2014 at 14:20

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