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ZFC works as a foundation because it can prove many sentences that are "translations" of theorems from "standard" mathematics into the language of ZFC.

But there's a subtlety. When we say, "ZFC can found most of mathematics," what do we really mean?

Do we mean

  1. ZFC proves most theorems (suitably translated into the language of sets) in the mathematics literature
  2. A metatheory that makes sense of "consistency" + the assumption that ZFC is consistent can be used to prove most theorems (suitably translated into the language of sets) in the mathematics literature
  3. A metatheory that makes sense of "models" + the assumption that ZFC has a model can be used to prove most theorems (suitably translated into the language of sets) in the mathematics literature
  4. A metatheory that makes sense of "models" + the assumption that ZFC has a standard model can be used to prove most theorems (suitably translated into the language of sets) in the mathematics literature
  5. Something else?
goblin GONE
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3 Answers3

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We mean that we can formalize the needed languages, and theories, and we can prove the existence of sufficient models for "regular mathematics" from ZFC.

This means that within any model of ZFC we can show that there are sets which we can interpret as the sets for "regular mathematics". Sets like the real numbers with their order, addition, and so on. And all the statements that you have learned in calculus about the real numbers and continuous functions, and so on -- all these can be made into sets which represent them and we can write proofs (which are other sets) and prove that these proofs are valid, and so on. All this within sets, using nothing but $\in$.

The key issue is that all this happens internally, so it happens within every model of ZFC, or rather in every universe (even if it is not assumed to be a set in a larger universe on its own).

Asaf Karagila
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    I can't quite work out whether you're saying (1) or (3). – goblin GONE Feb 27 '13 at 03:29
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    First note that (2) and (3) are really the same thing. If ZFC is consistent then it has a model (although not necessarily standard). But I am saying that (1) is sufficient. We don't need to add Con(ZFC) to our theory, even without contradictions we can prove a lot. Of course that if ZFC is inconsistent we can prove more; but we still exhibit proofs for the rest of the things without the need for the contradiction. – Asaf Karagila Feb 27 '13 at 03:31
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    Fair enough, but doesn't this contradict the mention of models of ZFC in your answer? – goblin GONE Feb 27 '13 at 03:34
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    @user18921: No. Why would it? Suppose that the collection of all sets is an actual concrete concept, and it happened to satisfy the axioms of ZFC -- but it doesn't have any set models of ZFC as elements. No set can be made into a model of ZFC. Sorry, it turns out that ZFC is "inconsistent" from the point of view of first-order logic. Within that universe we can still construct "regular mathematics". – Asaf Karagila Feb 27 '13 at 03:36
  • Now suppose that you have a model of ZFC, then you have a model of ZFC in which does not recognize any set within itself to be a model of ZFC as well. This model is just our universe from the previous comment, if we had to assume existence of models for ZFC then in that universe we couldn't have built mathematics; but we can. – Asaf Karagila Feb 27 '13 at 03:38
  • I'm afraid I don't really get it. Anyway, I edited the question to try to make it clearer. – goblin GONE Feb 27 '13 at 03:41
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    @user18921: I'm afraid that to your edit I will leave my answer unchanged. But look, it's an extremely delicate point (and many people don't care about it as well). It takes a lot of time to understand it, and I doubt you can truly grok this idea without spending a year or two doing set theory and understanding the delicate points about models of ZFC and internal vs. external, and how to encode things as sets. It's hard work, but then you look back and you understand it all. – Asaf Karagila Feb 27 '13 at 03:44
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    I will continue with a confession that a lot of what I wrote two years ago on this site are things I am embarrassed to read. I also read many answers on MO many times, and tried to understand and convinced myself that I do in order to continue. But only after working a lot with hands on set theory I finally had that understanding on how things can be modeled within ZFC. It's not something you can read off a book, or hear someone explain to you, and really understand it. You have to work very hard to earn this understanding if you really want it. – Asaf Karagila Feb 27 '13 at 03:46
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    To be honest, I really just want a "workable" understanding of how mathematics "fits together." Such an understanding seems to be very hard to come by, however. Edit: And I don't know if I want to spend two years learning set theory just to finally understand something that in the end I'm not even able to communicate. – goblin GONE Feb 27 '13 at 03:53
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    @user18921: Yes, that would be extremely difficult because modern mathematics is very fragmented. There are many fields which are very very distant from one another, and people in one don't know anything about the other. One of the professors in classical analysis asked me a really silly set theory question the other day; and I have no idea how to approach a problem in PDE. This is how modern mathematics is, and you have to pick your position. With time you gain the needed understandings, or improve your ability to learn so you can gain them on your own. – Asaf Karagila Feb 27 '13 at 03:55
  • Okay but lets be less ambitious. Suppose I want to understand not all of modern mathematics, but merely how to write a mathematical work that is very, very rigorous. It seems like even this is beyond my reach. Clearly, this is a little frustrating. (not that i'm saying its your fault or anything). EDIT: The problem is, I don't understand how to "use" a foundations. – goblin GONE Feb 27 '13 at 03:58
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    Well, rigor has little to do with set theory, ZFC, categories or otherwise mathematical foundations. If one is writing a rigorous proof about Lie algebras, they don't use ZFC, they don't care for it either (most time, anyway). They have a language which includes the operations and symbols needed and they have axioms for what is a Lie algebra, and they infer the conclusions they want to prove. That is all. – Asaf Karagila Feb 27 '13 at 04:03
  • To give an example from group theory: we're always saying things like "Let $C_2$ denote a cyclic group of order 2." How do we even know such a thing exists? It needs to be proved. But the group axioms do not prove the existence of $C_2$. Edit: Or maybe they do, but they do not prove the existence of infinite groups. Anyway, the point is that the group axioms are insufficient. – goblin GONE Feb 27 '13 at 04:05
  • Yes, which is what I meant when I said that ZFC allows us to prove the existence of "enough" models for mathematical theories. But the key point is that people work under the assumption their theory is inconsistent. You assume you work with consistent theories, until the point you derive contradiction and then you know that somewhere along the way something bad happened, and you go find that bad thing. Occasionally it was your own assumptions that are faulty, that is the theory is inconsistent. But you don't work with this in mind. – Asaf Karagila Feb 27 '13 at 04:13
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    Furthermore many people start from some assumptions, e.g. the natural numbers and work from there. So it suffices to show ZFC has a model of the natural numbers. And it does. So the rest follows. Or the real numbers, from which we can define vector spaces and measures and so on. Once you prove they exist, it is enough. And so on and so forth. He set theory used by most mathematicians is naive and easily interpreted in ZFC (unions, power sets, products, and so on). So we tied everything in a neat little bow... – Asaf Karagila Feb 27 '13 at 04:16
  • Okay I'm beginning to get it. But clarify this for me. Suppose we have that "the natural numbers have a model", and we want to use this prove the existence of new groups, for example. How does one use the statement that "the natural numbers have a model" to prove the existence of new groups? And does this not require some axioms (e.g. those of ZFC)? – goblin GONE Feb 27 '13 at 04:26
  • How much of the answer assumes things about consistency, or models, or a standard model, of ZFC? It seems to me that to answer the question you need to treat ZFC from an external viewpoint that does not assume any of these things, but describes its relation to non-formalized mathematics. – zyx Feb 27 '13 at 06:40
  • @user18921: Using completely naive set theory we can construct and define $\Bbb Z$ from $\Bbb N$. We can go on to define its quotients, and direct sums and products; if we have $\Bbb N$ then we essentially have all the finite groups whatever they are. And so on. This is just naive set theory (products exist, power sets exist, stuff like that) which is easily implementable in ZFC. But we don't need a model of ZFC for all that. Because a model of ZFC is a set. And it is consistent with ZFC that there are no such sets. So what would happen then? – Asaf Karagila Feb 27 '13 at 12:56
  • If you insist that there is a model, or a standard model, or whatever, of ZFC then you have effectively insisted that in models of ZFC where there are no models of ZFC inside you cannot do mathematics. This is preposterous. We care that we can formulated the naive set theoretical arguments within ZFC, and that way whenever we know that the things which are sets conforms to the rules of ZFC we know we can formalize mathematics within them. That's all. Whether or not ZFC itself is consistent is another question. – Asaf Karagila Feb 27 '13 at 12:59
  • @zyx: Read my previous comments. – Asaf Karagila Feb 27 '13 at 13:25
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    @Asaf: I agree that we can imagine a world where there is no set model of ZFC but the collection of sets happens to satisfy ZFC. However, that would be a world that contains a formal proof of a contradiction from ZFC. Since (by assumption) ZFC is true (about this hypothetical world if about nothing else), this means the the Gödel number of that formal proof would be something that doesn't belong to the true naturals -- so I'd say that cannot actually be our world. (On the other hand, our world could be one where the true sets don't satisfy ZF). – hmakholm left over Monica Feb 27 '13 at 15:10
  • (Hm, actually I'm not sure that my comment above was a response to anything you wrote here. I thought it was when I wrote it, but now I cannot find it. It may have been all in my head). – hmakholm left over Monica Feb 27 '13 at 15:15
  • @Asaf, the answer (paragraph 2) speaks of "within any model of ZFC" and talks about models and what is true "internally". Even granting the completeness theorem for FOL, it seems to me that you are giving a semantic answer while the question is essentially syntactic. We are talking about a formal system considering it as a formal language for encoding an informal language. – zyx Feb 27 '13 at 15:18
  • Another thing is that the question assumes that an adequate translation into the language of sets is available, so there is no need to explain that real analysis, number theory, geometry etc can be done in this framework. The OP is asking only about whether additional metamathematical principles of consistency or model existence are part of the translation. The answer to this question is the simplest possible one: if it was not in the informal mathematical proofs, it will not be in the translation to a formal proof system. – zyx Feb 27 '13 at 16:02
  • Separate from the preceding comments, I would like to understand the words "we can write proofs (which are other sets) and prove that these proofs are valid, and so on". In my understanding, proofs are syntactic objects in a formal language, such as ZFC, and if you were to fix some encoding of those objects as ZFC-sets, ZFC would not be able to prove much about the validity of that proof system, though it could check the syntactic correctness of individual proofs and some infinite families of proofs. – zyx Feb 27 '13 at 16:15
  • @Henning: I agree that it wouldn't be a satisfactory philosophical foundation. It's rather whimsical to think that T+$\lnot$Con(T) can be used as a foundational theory for mathematics, but it can. I don't share this particular philosophical point of view (as I recently debated with myself), and I realized that the thing I find important is actually LEM. If not for everything then for some things. But that really have nothing to do with this already convoluted comments thread... I'll stop now. :-) – Asaf Karagila Feb 27 '13 at 16:21
  • @zyx: Feel free to add this point to your answer. When I think about foundation for mathematics, I think about it in a semantical way being reassured by the completeness theorem. But ZFC can be used to develop syntactical consistency this way as well. (1) Write the proof for the completeness theorem; (2) write the proof that such and such language and such and such structure is a model for such and such theory; (3) prove that every model of that theory satisfy a certain theorem; (4) conclude that the theory proves the sentence from the above. – Asaf Karagila Feb 27 '13 at 16:24
  • @zyx: I'm not sure what you are asking about in the second comment. – Asaf Karagila Feb 27 '13 at 16:28
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    @Henning: It may not have been a response to Asaf, but it did answer a question I hadn't asked, namely to get a second opinion on my conclusion that a model of ZFC containing no (internal) models of ZFC must have nonstandard naturals. –  Feb 27 '13 at 18:07
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    @Hurkyl: This was mentioned in quite a few answers on the site, actually. The point is that if your universe is such universe then you really can only say that one has to discern between the natural numbers in the meta-theory and the natural numbers in the model. Of course this is a reasonable meta-mathematical cause to reject such universe as the meta-theory to begin with; but this is not a mathematical reason per se. – Asaf Karagila Feb 27 '13 at 18:10
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I don't think it has anything to do with models of ZFC, so I'd say (1) is the closest. Of course it's not the sheer number of statements ZFC proves that is important, it's the fact that among those statements one can find, for almost any mathematical theorem, a formal theorem of ZFC that captures the meaning of that theorem (albeit in a very pedantic way.)

The main exception to this is theorems of set theory itself, which may require something like ZFC + large cardinal axioms to prove.

Trevor Wilson
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Only (1). Unformalized mathematics does not refer to its own consistency or existence of its own models, and its translation into a system like ZFC would not, either.

zyx
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