0

The proof of the independence of $\mathsf{CH}$, goes like this: one proves in $\mathsf{ZFC}$ that assuming $\mathsf{CON(ZFC)}$ there is a model of $\mathsf{ZFC}$ in which $\mathsf{CH}$ holds and another one in which it fails. This is well and good, but it makes me wonder how exactly does it translate to the meta-theory(say $\mathsf{PA}$), how could we conclude that $$\mathsf{metatheory}\vdash \mathsf{CON(ZFC)}\to \mathsf{ZFC}\not\vdash \mathsf{CH}?$$ After all, this is what we mean when we say that there is no mathematical proof of the continuum hypothesis. The problem is that the encodings are all different.

Vivaan Daga
  • 5,531
  • 3
    "this is what we mean when we say that there is no mathematical proof of the continuum hypothesis" - really? When I make a mathematical assertion, like "all groups of prime order are cyclic", I'm not talking about what can be proved in a weak metatheory like PA. I'm talking about what I can prove using the whole apparatus of mathematics. The situation is no different for mathematical assertions about logic, e.g. the independence of CH from ZFC. – Alex Kruckman Jun 24 '22 at 15:15
  • 2
    Of course it can be very interesting to study what theorems can be formulated and proved if we tie our hands behind our backs and adopt a weak arithmetic metatheory. That question, in the specific case of independence results has been asked before (e.g. here and surely many other times). But this is certainly not what people mean by default! – Alex Kruckman Jun 24 '22 at 15:17
  • @AlexKruckman So you take ZFC for granted? Taking ZFC as your meta-theory. When you prove that all groups of prime order are cyclic, you give a ZFC proof in PA. – Vivaan Daga Jun 24 '22 at 15:18
  • 2
    I have no idea what it means to "give a ZFC proof in PA". When I prove that all groups of prime order are cyclic, I write a proof in natural language that I am confident could, if one wanted to spend the effort, be formalized in ZFC. I think this is the most accurate description of the default meta-theory in use by mathematicians: natural language proofs that could, in principle, be formalized in ZFC. – Alex Kruckman Jun 24 '22 at 15:21
  • And what is the meta theory you give your proof in? ZFC? One of the goals of Hilberts program was to convert all math into finitary reasoning only possible if you have a weak meta theory @AlexKruckman – Vivaan Daga Jun 24 '22 at 15:23
  • 1
    In my last comment, I described my metatheory (which is also, I assert, the default metatheory for the vast majority of mathematicians). See Asaf's answer here for some nice discussion on metatheories for set theory. – Alex Kruckman Jun 24 '22 at 15:27
  • 2
    I think you have to accept that different people are going to use different metatheories, @OP. However, this question is independently interesting (so in my opinion the last two sentences should be removed since they're not actually needed). That said, what isn't already answered in this earlier question? – Noah Schweber Jun 24 '22 at 15:27
  • @NoahSchweber I closed the question as a duplicate, and the conservativity argument in your answer to the duplicate question is very nice. But it would be nice to also link to a reference that gives some hint as to how a translation into syntactic manipulations of proofs would go (as suggested in Andreas Blass's answer). Do you have a good reference in mind for this? – Alex Kruckman Jun 24 '22 at 15:32
  • 1
    @AlexKruckman I’m interested in your philosophy so if you don’t mind me asking: You take ZFC as your theory where you prove things, you also take ZFC as your meta theory and meta meta theory. Then why would you care about say the consistency of ZFC? You already take it for granted, no? – Vivaan Daga Jun 24 '22 at 15:37
  • @NoahSchweber Can every independence proof be translated by your method given in the other question? – Vivaan Daga Jun 24 '22 at 15:37
  • I don't understand the distinction between "theory where you prove things" and "meta theory", much less "meta meta theory". See my answer here. Yes, we can say my meta theory is ZFC. That's the theory where I prove things. Now I can consider various logics and theories (object theories) and try to prove things about them, and what they prove and don't prove. One of these object theories I might consider is ZFC, or rather the internalized version of ZFC, whose axioms match those I use at the meta-level. – Alex Kruckman Jun 24 '22 at 15:44
  • Consistency of my meta theory is not something I "take for granted", but it is something that I really strongly hope is true, otherwise I'm wasting my time. Unfortunately, it's a question that can only be settled negatively by find a proof of a contradiction. Consistency of my object theory is an interesting mathematical question to consider, which I could hope to settle one way or another - except of course Gödel showed that when the object theory is ZFC, I don't have much hope of settling the question positively! – Alex Kruckman Jun 24 '22 at 15:48
  • @VioletFlame "Every independence proof" is far too broad to address. However, the method I describe there does work for every forcing-based independence proof, as well as many more. – Noah Schweber Jun 24 '22 at 15:50
  • @AlexKruckman: It's a bit vague, but I don't think inconsistency of ZFC or whatever would necessarily mean you're wasting your time. Even if ZFC turned out to be inconsistent, your description of the way we write mathematics as "natural language proofs that could, in principle, be formalized in ZFC", is IMHO more accurately described as "natural language proofs that could, in principle, be formalized in any reasonable foundational metatheory". I think good math does not depend that much on the metatheory. – tomasz Jun 24 '22 at 16:44
  • @tomasz Oh yes, I totally agree with that. – Alex Kruckman Jun 24 '22 at 16:45
  • 1
    Trying to straighten out the confusion about "meta": In mathematics generally, we use certain basic principles that (we know by experience) can be formalized in an axiomatic theory, usually ZFC. In branches of math other than logic, this theory is the only theory involved, and we usually don't mention it. In logic, though, theories are a part of the topic being studied, and they may be very different from the theory we're using. So people invented the name "meta-theory" for the theory we're using. The rest of mathematics uses the same theory in the same way, but has no need to call it "meta". – Andreas Blass Jun 24 '22 at 20:02
  • @AlexKruckman In https://mathoverflow.net/questions/248965/do-set-theorists-use-informal-set-theory-as-their-meta-theory-when-talking-about JDH states that PA suffices as a meta-theory and is what set theorists use. – Vivaan Daga Jun 25 '22 at 05:13
  • @AlexKruckman Genrally speaking one takes the meta-theory for granted,would you accept a proof which uses infinitary reasoning in order to show "ZF proves X"? – Vivaan Daga Jun 25 '22 at 08:11
  • @AndreasBlass In order to prove something about ZF, one would want to be able to prove it in a finitistic system, so everyone can (even the fintists) would accept the result. No? – Vivaan Daga Jun 25 '22 at 08:15
  • 1
    In the comments to the answer you linked to, JDH writes "while I recognize that a weak meta-theory suffices for all the usual arguments, my own practice is not particularly focussed on a weak meta-theory. I feel free to assume whatever it might take in the meta-theory to push an argument through." There is a difference between recognizing that a weak meta-theory would suffice and actually using it! – Alex Kruckman Jun 25 '22 at 11:35
  • "would you accept a proof which uses infinitary reasoning in order to show "ZF proves X"?" Whether I accept a proof does not depend on the form of the statement being proved. I have the same standard of proof for, say, the intermediate value theorem, as I do for "ZF proves X". I think it would be very strange to adopt different standards in different contexts. Now let's say you give me a proof (in ZFC) of "ZF proves X". I would accept this as a proof of "ZF proves X", but that's slightly different than a proof of X. I would much rather have in hand a proof of X. – Alex Kruckman Jun 25 '22 at 11:48
  • @AlexKruckman So you do not care about devloping ZFC in a fintistic system so even the finittists accept it, atleast formally(Hilbert wanted this). Correct? Thats interesting. But i'd like to point out that not all logicians have this view for instance see Kunenvset theory book, esp the appendix on meta-math or the chapter on forcing, where he makes a point to show things finitistically. – Vivaan Daga Jun 25 '22 at 14:01
  • 3
    If I were writing textbook, I'd probably mention that certain things can be proved finitistically. But in my ordinary work, I don't worry about that. Like @AlexKruckman, I use the same tools for my work in logic and set theory that I use in any other part of mathematics. – Andreas Blass Jun 25 '22 at 14:09
  • @AndreasBlass Hilbert desired a finitary consistency proof of ZFC, and wanted that any deductions made in ZFC could be accepted by finitists are you saying modern logicans do not care too much about finitists? – Vivaan Daga Jun 26 '22 at 06:58
  • @AlexKruckman Id like to to adress the above comment to you as well. – Vivaan Daga Jun 26 '22 at 06:58
  • 1
    Hilbert desired a finitary consistency proof for strong systems (at least second-order arithmetic, perhaps even ZFC), but Gödel's second incompleteness theorem implies that there is no such proof, not even for first-order arithmetic. Modern logicians, in particular proof theorists, care about what must be added to finitist principles in order to get such consistency proofs --- for example well-ordering of larger ordinals (following Gentzen) or primitive recursive functionals of higher types (following Gödel) --- but there are other branches of modern logic not concerned with this. – Andreas Blass Jun 26 '22 at 11:36
  • @AlexKruckman Last question regarding one of your comments above suppose we have $\mathsf{ZFC}\vdash(\mathsf{ZFC}\vdash X)$ can we conclude $\mathsf{ZFC}\vdash X$? – Vivaan Daga Jul 01 '22 at 07:39
  • Why not ask a new question? – Alex Kruckman Jul 01 '22 at 09:00
  • @AlexKruckman I didnt think it’s worthy enough to be a new question(it probably has a trivial answer), since you mentioned something like this in an above comment I asked you. Is the answer not trivial? – Vivaan Daga Jul 01 '22 at 09:32
  • 1
    Well, the point is that the comments are not supposed to be used for extended discussion like this. I'm happy to clarify my personal philosophical positions, since these are not on topic for the main site, but now you're asking a concrete mathematical question. The answer is "not without extra (very reasonable) assumptions". But its a subtle enough point that it would make a good question. – Alex Kruckman Jul 01 '22 at 13:26

0 Answers0