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I have learned that ZFC has not been proven consistent, and that further more if one were to start from ZFC and prove ZFC consistent, this would imply that ZFC is not consistent, due to Gödel.

A few questions about what I just said: Does this mean there is a possibility that ZFC is inconsistent? Are there some really good (mathematical) arguments for why, even though ZFC might be inconsistent, it probably is consistent? (or at least that we should still use it even if we aren't sure it's consistent?) What would happen if somebody were to prove that ZFC were inconsistent? Would we have to throw out all of our old proofs? And lastly, does Gödel's work imply that, in order to prove ZFC consistent (without automatically proving ZFC inconsistent in the same stroke) we would need to make more assumptions outside of ZFC? In other words, is there hypothetically a way of proving ZFC consistent without making more assumptions?

I would appreciate answers to even one of these questions.

Perry Bleiberg
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  • Answer to the first question: Despite many people spending years on looking for inconsistency, no one has found any. – Wojowu Jul 05 '15 at 16:51
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    The sun would still rise tomorrow, and the birds sing. Most of mathematics would remain unaffected. The rest would require minor tweaking. – André Nicolas Jul 05 '15 at 16:54
  • @Meelo Thank you for sharing – Perry Bleiberg Jul 05 '15 at 16:58
  • @André: What if the contradiction would actually be discovered in finitary parts of set theory. Namely, it would show that $\sf PA$ is inconsistent as well? Would most mathematics remain unaffected at that point as well? – Asaf Karagila Jul 05 '15 at 17:05
  • With PA here would be a problem. Luckily, it is consistent, else there would not be a universe. – André Nicolas Jul 05 '15 at 17:11
  • @AndréNicolas: I don't see how the universe has anything to do with PA. PA is a good approximation, sure, but the claim that it underpins the universe seems to have no basis. Otherwise, you should at least be able to point to a real-world interpretation of PA that fits with current science. – user21820 Sep 13 '18 at 14:15

2 Answers2

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First of all, proofs don't exist in vacuum.

We cannot prove the consistency of $\sf ZFC$ from theories like $\sf ZFC$ itself, or even $\sf PA$. But we can prove the consistency of $\sf ZFC$ from other theories, stronger theories. For example, from $\sf ZFC+I$ where $\sf I$ is the axiom stating that there exists an inaccessible cardinal we can prove that $\sf ZFC$ is consistent.

This is an answer to your last question. To prove $\sf ZFC$ is consistent with need to work in theories which are not "just $\sf ZFC$ itself".

Now. Is it possible that $\sf ZFC$ is inconsistent? Yes. It is possible. What happens if it is inconsistent? No bridges will collapse, that's for sure. We'll investigate the inconsistency to see what caused it. After we've understood that, we'll try to rescue whatever we can from the mathematics of the last 200 years, and we'll proceed as before, pushing mathematics to the limit.

And as for the arguments that $\sf ZFC$ is consistent. Well, self-evidency for one. I think that a lot of the axioms are quite natural. Perhaps the power set axiom is a bit too much, but the rest of the axioms are really quite natural and "un-intruding", in the sense that you don't feel them when you do your work. Which is a good thing from a foundational theory. Another argument is that we haven't found any contradictions so far, and we've been pushing nearly a century since $\sf ZFC$ was established. Some very smart people looked into that, and if they haven't found any, there's a good chance we won't find that contradiction either.

Both these arguments are a bit silly, and a bit circular or fallacious. There are no "good" arguments. This is a matter of belief, if you want to believe that $\sf ZFC$ is inconsistent, then by all means find a better alternative. The rest of us will continue to do mathematics as we did until now.

Asaf Karagila
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  • I would personally say that axiom of regularity is one axiom which one could argue whether it's natural or not. If it weren't for me sitting inside $V$ for few years now, I'd probably argue that set containing itself is just as natural as set containing no elements. – Wojowu Jul 05 '15 at 17:25
  • I agree that regularity is a bit of an oddball, but (1) it does not add new contradictions; (2) regularity is really equivalent to $\in$-induction, which makes a lot of sense. – Asaf Karagila Jul 05 '15 at 17:28
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There are two separate questions here:

  • What impact would an inconsistency in ZFC have on mathematics?

  • What ways do we have of concluding that ZFC is consistent?

For the former, see What if the current foundations of mathematics are inconsistent? and What would be some major consequences of the inconsistency of ZFC?; I don't think I have much to add here. The short version is: mathematics tends to use much less than ZFC, and the vast majority of math wouldn't be directly effected by a proof that ZFC is inconsistent.

For the latter, here's the situation (for simplicity, I'll assume ZFC is consistent):

  • Godel's work shows that ZFC cannot prove "Con(ZFC)," where Con(ZFC) is a sentence in the language of set theory which, pretty clearly, states that ZFC is consistent. There are people who argue about whether Con(ZFC) actually expresses "ZFC is consistent;" for now, though, I'll take this for granted.

  • This means that any theory T which proves Con(ZFC) must have axioms which ZFC doesn't. This doesn't really mean that T must be stronger than ZFC, just that ZFC is not stronger than T. In principle, each of T and ZFC could prove things the other couldn't.

  • This leads to the following program: fix a very weak base theory B (in particular, substantially weaker than Peano arithmetic), and ask: "What axioms do we need to add to B to prove Con(ZFC)?" Note that insofar as these axioms are intuitively clear, this would provide evidence that ZFC is in fact consistent, so this program can be used as an attempt to intuitively show the consistency of ZFC.

  • It turns out this can be done! Very very roughly, given any reasonable theory $T$ (including ZFC), there is some ordinal $\alpha$ such that B+"$\alpha$ is well-founded" can prove the consistency of $T$. The bigger the $\alpha$, the "harder" it is to prove that $T$ is consistent. This $\alpha$ is called the proof-theoretic ordinal of $T$, and has been extensively studied, and there are many mathoverflow questions about it; see https://www1.maths.leeds.ac.uk/~rathjen/ICMend.pdf.

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Noah Schweber
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  • To check my understanding: Is the proof-theoretic ordinal of ZFC what others have called "the inaccessible ordinal"? I see @Asaf Karagila referred to an "inaccessible cardinal" in his answer. Is his mention of ZFC + I a specific case of what you're talking about here? – Perry Bleiberg Jul 05 '15 at 18:41
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    No, an inaccessible cardinal is a very different sort of animal. In particular, proof-theoretic ordinals are countable (and, in fact, recursive!). – Noah Schweber Jul 05 '15 at 21:58