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Would "There is no greatest integer" be sufficient axiom to add to Peano to make the strengthened finite Ramsey theorem provable? If so, what example can be offered of an unprovable statement in the new theory? Instinct tells me there might not be one, but then I am quite naive.

What I like about this axiom is that it adds nothing, it only takes away, which is what any axiom that creates something that doesn't really exist must do.

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    "For every $x$, $x+1$ exists" is already a Peano axiom. Am I missing something? – Arthur Aug 15 '16 at 13:31
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    As long as your system is recursively axiomatizable and consistent, Godel's incompleteness theorem will apply. – James Aug 15 '16 at 13:31
  • @Arthur no, almost certainly I am missing something. Then I do not understand what the axiom of infinity would add. I suspect I should delete rather than take the cascade of downvotes! – it's a hire car baby Aug 15 '16 at 13:32
  • What do you mean by 'strengthened finite Ramsey theorem"? A typical minimal formulation of the Peano axioms does not provide an inequality relation. – hardmath Aug 15 '16 at 13:34
  • To extend Arthur's comment, suppose you have a set $S$ and a unary function $f:S\rightarrow S$. Suppose $f$ is injective and there is some $x\in S$ which is not in the image of $f$. You can show $S$ must be infinite. The axioms of PA imply that $x\mapsto x+1$ is such a function. – James Aug 15 '16 at 13:34
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    The axiom of infinity is an axiom about sets, not about integers. As long as the natural numbers satisfy the axioms of PA, there is no "reasonable addition". And certainly nothing that would "explode" into a full universe of set theory. – Asaf Karagila Aug 15 '16 at 13:55
  • @hardmath good point! I read that it was one of the first common-sense statements expressible in the language of arithmetic but not provable in Peano. But reading it, it's full of set theory language, so presumably it can't even be expressed in the language of Peano? – it's a hire car baby Aug 15 '16 at 16:50
  • @AsafKaragila How about something like $\lvert\mathbb{N} \rvert =a: n\leq a\forall n\in\mathbb{N}$. That gives you an arbitrarily large number $a$ which is equal in magnitude to the count of the natural numbers. – it's a hire car baby Aug 15 '16 at 17:00
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    There is no model of Peano where $\Bbb N$ is even definable, except the standard model, and if it were, that would violate the induction schema. Moreover, you can't really talk about sets in Peano, sure for a limited time offer you could talk about definable sets by using truth predicate to ask whether or not a certain numeral satisfies a certain formula (which defines a set). But not much more than that. So if you go to non-standard models, you lose a lot of that power. So in either case, this is a "no go". – Asaf Karagila Aug 15 '16 at 17:46
  • I intended my axiom to define $\lvert\mathbb{N} \rvert $, or we could equally call it $\infty$, I just chose a recognisable symbol. It doesn't have to equal the $\lvert\mathbb{N} \rvert $ we know. Or we could just leave it at $a$. Is that still a no no? – it's a hire car baby Aug 15 '16 at 18:05
  • @RobertFrost If you take a nonstandard model of PA, then you can have nonstandard integers; you can pick one of them and announce "this is the size of the naturals!" if you like, but you then have to be very clear exactly what you mean by that. – Patrick Stevens Aug 16 '16 at 10:38
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    Not sure this will help with your Ramsey theorem, but a set $X$ is said to be Dedekind-infinite if and only if there exists $f: X\to X$ such that $f$ is injective, but not surjective. And the usual successor function on the set of natural numbers is just such a function. So, postulating Peano's Axioms in itself amounts to postulating the existence of an infinite set. Taken as a whole, they could be seen as an axiom of infinity. – Dan Christensen Sep 01 '16 at 19:01

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"There is no greatest (non-negative) integer." trivially translates to the sentence "$\neg \exists m\ \forall n\ ( n \le m )$" over PA, where "$x \le y$" is short-form for "$\exists t\ ( x+t = y )$". This sentence is easily provable over PA, which I shall leave as an exercise.

Consequently, adding your proposed axiom to PA does nothing at all, and all unprovable sentences over PA remain unprovable over the new system. To even state Ramsey-type theorems in PA one has to go through some encoding, which is not a trivial matter, and it is inadvisable to attempt to understand that without a firm foundation in logic. I'm serious; running before being able to walk is dangerous. I shall yet again encourage you to spend a few months working through proper logic texts such as the first two listed in this post.

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user21820
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  • Thanks, very clear. What about the alternative proposed axioms below, that define the size of the integers by either a metric (not prefered) or a well-ordering relation (which I'm more attracted to)? Can they add anything to Peano? I do see the difficulty in defining $\geq$. – it's a hire car baby Aug 16 '16 at 19:16
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Presumably "strengthened finite Ramsey theorem" means the Paris-Harrington theorem; if not, what does it mean?

Peano arithmetic already proves "There is no largest natural number," so adding this statement to PA just yields PA again.

If you're interested in the axioms you need to add to (say) PA in order to prove certain arithmetic facts, you should look up reverse mathematics. But this is a fairly advanced subject, and you should be familiar with the basics of proof and model theory before you tackle it. Your question suggests you are not quite comfortable with the model theory of first-order logic, so I suggest you start there (Ebbinghaus-Flum-Thomas is a good source; Marker is better written and more advanced, but has typoes).

Meanwhile, (Rosser's strengthening of) Goedel's theorem constructively provides a sentence $\varphi$ which is undecidable in a theory $T$, whenever $T$ is computably axiomatizable, consistent, and contains (say) Robinson's $Q$. So I don't understand the last two sentences of your first paragraph.

EDIT: See also Why is the Axiom of Infinity necessary?.

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Noah Schweber
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  • Thanks v. helpful. I like the sound of "reverse mathematics"! The "strengthened finite Ramsey theorem" is the theorem which the Paris-Harrington theorem proves is unprovable in Peano. (Apparently). – it's a hire car baby Aug 16 '16 at 19:20
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    @RobertFrost Based on the end of your comment, let me explain very briefly how the proof that Paris-Harrington goes: first, it is shown that the relevant Ramsey-type statement implies (over PA) that a certain linear order, $\epsilon_0$, is well-founded. (This isn't actually totally accurate, but it's reasonably close to true; since PA is a first-order theory, we have to be careful how we talk about structures inside PA.) Separately, Gentzen showed that induction along $\epsilon_0$ implies (over a subtheory of PA) the consistency of PA. So by Godel's 2nd, we know PA doesn't prove the theorem! – Noah Schweber Aug 17 '16 at 01:32
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    Look up proof-theoretic ordinals for more information. – Noah Schweber Aug 17 '16 at 01:32
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    @RobertFrost: As Noah mentioned, you will need basics in both proof theory and model theory to be able to understand reverse mathematics, otherwise it will be utterly confusing. (Indeed I made neither head nor tail of it before I learnt enough proof theory and model theory!) Note that what he calls basic are in fact what I've suggested you learn first, plus some more. Specifically, Stephen Simpson's notes are introductory material for proof theory, while Rautenberg's book has a chapter on introductory model theory (simpler than Marker so you might want to read Rautenberg first). – user21820 Aug 17 '16 at 08:30