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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?
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3 Answers 3

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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).

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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Commented Feb 27, 2013 at 3:31
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    $\begingroup$ @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". $\endgroup$
    – Asaf Karagila
    Commented Feb 27, 2013 at 3:36
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Commented Feb 27, 2013 at 3:46
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    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Commented Feb 27, 2013 at 3:55
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Commented Feb 27, 2013 at 4:03
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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.

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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.

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