EDIT: I am not asking about the validity of exponentiation, or PA. My question is about a specific technical claim which Nelson makes in this article (pp. 9-12): that a certain theory does not prove a certain sentence, and more generally that that theory does not prove any of a certain class of sentences. I am not interested in the mathematical quality, philosophical validity, literary value, font choice, overall moral rectitude, or shoe size of the article as a whole. I hope my edits have clarified this, and it is now apparent that the philosophical context of this question is merely that: context.

Let's suppose we are skeptical that PA is in fact consistent; perhaps we are convinced that addition "makes sense," pretty sure multiplication "makes sense," but dubious that exponentiation "makes sense" (this appears to not be too far from Nelson's own opinions, based on the article linked above). It now becomes valuable to have a notion of relative finitism: if we accept that one operation makes sense finitistically, what other operations can we argue are acceptable on that basis alone?

Informally, we want to ask:

Given definable functions $f, g$ (growthwise in the neighborhood of the "usual arithmetic functions"), can we prove that, if there is a "natural number series" closed under $f$, then there is also a "natural number series" closed under $g$?

Of course, the word "prove" is a dangerous one there: if we mean prove in PA, we're trivializing everything right from the outset even if we're confident PA is consistent. On the other hand, replacing PA with a weaker theory seems to beg the question of how to justify the finitistic acceptability of that theory.

Nelson suggests the following approach: start with PA, but somehow modify it so that it can imagine proper initial segments of the universe which are closed under successor. Now we can ask nontrivial questions about the existence of "well-behaved initial segments" - intuitively, "notions of number" that permit the operations we care about to make sense - and we can do so from the perspective of PA even without accepting PA!

Specifically, Nelson considers the theory PA', in the language of arithmetic + a new unary predicate symbol $C$ (counting number), consisting of PA together with the statement "$C$ is downwards-closed, contains $0$, and $\forall x(C(x)\implies C(x+1))$."

Although PA' contains PA, it is still extremely weak in a sense: since we haven't extended the induction scheme to include formulas involving $C$, PA' can't prove the "obvious" statement $\forall x(C(x))$," or even that $C$ is closed under addition! So we're in a very interesting situation: on the one hand, we have a lot of deductive power at our disposal from the "ambient PA-ness," but on the other hand we've also given ourselves tools for creating contexts in which arithmetic breaks very badly.

Nelson uses this as a platform for asking the above question in a rigorous form.

  • Claim 1: There is a definable initial segment of $C$ which PA' proves is closed under addition (and successor).

    • Proof: Let $A=\{x\in C: \forall y\in C(y+x\in C)\}$. Downwards closure and closure under successor are easy to prove. For closure under addition, note that if $x_1,x_2\in A$ and $y\in C$, we have $y+(x_1+x_2)=(y+x_1)+x_2$, and $y+x_1\in C$ since $x_1\in A$, so $(y+x_1)+x_2\in C$ since $x_2\in A$; that is, $x_1,x_2\in A\implies x_1+x_2\in A$.
  • Claim 2: There is a definable initial segment of $C$ which PA' proves is closed under multiplication (and addition and successor).

    • Proof: Let $M=\{x\in A: \forall y\in A(y\cdot x\in A)\}$. Downwards closure and closure under successor and addition are easy to prove. For closure under multiplication, note that if $x_1,x_2\in M$ and $y\in A$, we have $y\cdot (x_1\cdot x_2)=(y\cdot x_1)\cdot x_2$, and $y\cdot x_1\in A$ since $x_1\in M$, so $(y\cdot x_1)\cdot x_2\in A$ since $x_2\in M$; that is, $x_1,x_2\in M\implies x_1\cdot x_2\in M$.

Note that in each case we've used associativity (which is proved in PA for all numbers, not just those in $C$; this is how PA provides "useful context" for our finitistic concerns, the point being that "addition is associative" is clearly acceptable relative to the claim that addition makes sense in the first place). This breaks down for exponentiation, of course. Here Nelson makes two claims, one explicit and the other implicit.

The claim Nelson makes explicitly is:

Weak claim: PA' cannot prove that the set $E=\{x\in M: \forall y\in M(y^x\in M)\}$ is closed under exponentiation.

It seems, however, that his real point is that this is a fundamental obstacle, that in some sense the definition of $E$ above is the only "reasonable" candidate. In other words, I think the following stronger claim is implicit in Nelson's critique of arithmetic:

Strong claim: PA' cannot prove that there is a definable initial segment of $C$ closed under exponentiation. (More precisely: there is no formula $\varphi$ in the language of PA' such that PA' proves that $\varphi$ defines an initial segment of $C$ which is closed under exponentiation.)

My question is:

Question: Are these claims correct?

I'm specifically interested in the stronger claim, since that seems to be the more significant one and a positive answer would have plausible foundational value; however, the weaker claim is probably easier to analyze, and is also the only claim Nelson explicitly made.

Let me mention, for additional motivation, two possible "spin-off" questions which may be of interest:

  • First, we could replace PA with a different theory of arithmetic. This would have the effect of changing what arithmetic results we could use in establishing the existence of a definable cut below $C$ with certain closure properties. The arguments above only require the most basic bits of arithmetic, but conceivably a more complicated argument could require a nontrivial amount of induction. If indeed replacing PA with a different theory of arithmetic would change the situation, that would be really cool, even if the foundational significance is not obvious.

  • Second, we can "relativize" Nelson's construction. Say that a definable (in the language of PA) function $f$ is finitistic relative to another definable function $g$ if there is a formula $\varphi$ in the language of PA$_g$ - which is the theory consisting of PA together with a unary predicate symbol $G$, and axioms saying that $G$ names a downwards-closed set closed under successor and $g$ - which PA$_g$ proves defines a downwards-closed set closed under successor and $f$. Relative finitism seems potentially interesting (and possibly connected with bounded arithmetic), even from a non-finitistic point of view, especially if per the above bulletpoint the "ambient arithmetic" can meaningfully affect the situation.

Noah Schweber
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    When I pretend to be a finitist (which happens once in a *few* blue moons), I like to think that I work inside a model of PA where Con(PA) is false. In particular, inductive reasoning must be internal, rather than external, since second-order induction fails. I find that to be helpful. – Asaf Karagila Aug 16 '18 at 22:50
  • Surely PA together with the obvious extension of the induction scheme to embrace $X$ proves $\forall x\cdot x \in C$? So $C$ seems to be irrelevant. – Rob Arthan Aug 16 '18 at 23:36
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    Isn't Nelson's position usually described as "ultrafinitism"? I thought ordinary "finitists" were at least okay with something like PRA? – hmakholm left over Monica Aug 16 '18 at 23:40
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    @RobArthan Everything's first-order here, and we're *not* extending the induction scheme to include formulas in the new language. As a concrete example, any end extension of models of PA $M\subseteq_{end}N$ yields a model of PA,' with "arithmetic part" $N$ and $C$ interpreted as $M$. – Noah Schweber Aug 17 '18 at 05:36
  • I suppose that when you say Nelson claims "PA' cannot prove ...", then we should interpret that as a _relative_ unprovability claim: "If PA is consistent, then PA' cannot prove ..."? – hmakholm left over Monica Aug 17 '18 at 14:56
  • And you need to rewrite "closed under exponentiation" to at least something like "closed under exponentiation and successor (and addition and multiplication?)"; otherwise $\{0,1\}$ would work. – hmakholm left over Monica Aug 17 '18 at 14:58
  • I wouldn't worry about any "possible collapse of contemporary mathematics" just yet. Perhaps you have discovered some limitation of first-order PA? (I don't follow your argument, so I can't comment.) In any case, as you must know, if we allow set theory, exponentiation makes perfect sense, though there is the matter of $0^0$. – Dan Christensen Aug 17 '18 at 15:06
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    @DanChristensen I'm not asking about the collapse of mathematics. I'm confident in the consistency of PA and indeed much more. I'm asking a **specific model-theoretic question** motivated by Nelson's remarks: whether for a certain theory PA' (which is *not* PA!) there is a formula in the language of that theory which PA' proves defines an initial segment of (the part of the model named by) $C$ which is closed under successor and exponentiation. While Nelson provides the context, my question is not about the validity of his concerns. (cont'd) – Noah Schweber Aug 17 '18 at 15:33
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    Meanwhile, the statement "if we allow set theory, exponentiation makes perfect sense" is pointless: if we're concerned about exponentiation, why on **earth** would we have any confidence at all in set theory (as normally formulated)? Of course I know how to develop arithmetic inside set theory, but this isn't helpful to me (since it has nothing to do with my question) or to someone who is dubious of exponentiation (since from their point of view, set theory would be *even more dubious* and so the development of arithmetic inside set theory wouldn't increase confidence in arithmetic at all). – Noah Schweber Aug 17 '18 at 15:36
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    I've edited for clarity, but: at a certain point in that paper, Nelson made a specific technical claim which as far as I can tell he did not justify. I'm asking - independently of the rest of the paper - whether *that specific claim* (or its natural strengthening) is true. I believe the technical claim in question is of philosophically-independent interest, although obviously that's subjective. – Noah Schweber Aug 17 '18 at 15:39
  • I was puzzled that you seemed dubious about exponentiation on N. I'm glad to see that is not the case. – Dan Christensen Aug 17 '18 at 15:42
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    @ChristianBlatter I'm not sure what you're getting at. The deleted answer had *literally nothing* to do with my question. My question, again **is not whether Nelson's critique of arithmetic is correct**: it's whether a certain *specific technical claim* which Nelson makes *along the way* is justified. This question is wholly independent of one's philosophical stance; in particular, **as I say in the question I am totally confident of the consistency of PA and indeed much more**, so I don't know why you would think I'd delete an answer saying "exponentiation is fine" because I disagree with it! – Noah Schweber Aug 17 '18 at 16:13
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    @ChristianBlatter And actually an answer to a question in analysis or combinatorics which fails to actually address the question *would rightfully* be deleted. I'm not sure why you think otherwise. – Noah Schweber Aug 17 '18 at 16:21
  • Good question, bad paper. – anomaly Aug 17 '18 at 16:22
  • Suppose the Weak Claim to be written in the language of set theory. Is the Weak Claim a theorem of ZFC? That is, can ZFC prove there is a model of PA' in which E is not closed under exponentiation? How does Nelson justify the Weak Claim if not by means of a model? – DanielWainfleet Aug 19 '18 at 01:53
  • @DanielWainfleet Nelson doesn't justify the weak claim **at all**: no evidence for it is given in the paper. A ZFC (or similar) proof that the weak claim is true would be exactly what I'm asking for. That's precisely my question! – Noah Schweber Aug 19 '18 at 05:25

1 Answers1


Here is a proof of the weak claim. Let $X$ be any nonstandard model of PA with an initial segment $I$ containing all standard numbers which is closed under addition and multiplication but not exponentiation. Let $C$ be the set of $x\in X$ such that $x\leq n^i$ for some standard $n$ and some $i\in I$. Note that $C$ is closed under addition and multiplication: if $x\leq n^i$ and $y\leq m^j$, then $xy$ and $x+y$ are both at most $(m+n)^{i+j}$. So, we can take $X$ as a model of PA' with this $C$, and your set $M$ will be $C$.

Note also that $C$ is closed under exponentiation to elements of $I$, since if $x\leq n^i$ then $x^j\leq (n^i)^j=n^{ij}$. It follows that your set $E$ contains $I$. On the other hand, if $e\in E$, then in particular $2^e\in C$ so $2^e\leq n^i$ for some standard $n$ and some $i\in I$. But we have $(2^m)\geq n$ for some standard $m$, and so $n^i\leq 2^{mi}$ and so $e\leq mi$. Thus $e\in I$.

Thus your set $E$ for this model is just $I$. Since $I$ was chosen to not be closed under exponentiation, this proves the weak claim.

Eric Wofsey
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  • Belatedly, thanks! I've accepted this and asked about the strong claim [in a separate question](https://math.stackexchange.com/questions/2955417/exponentiation-and-a-weak-fragment-of-arithmetic). – Noah Schweber Oct 14 '18 at 18:19