# Check my proof of "overspill" in non-standard models of Peano arithmetic.

Proposition

Let $\mathcal{M}$ be a nonstandard model of Peano arithmetic, $\phi(v,\bar{w})$ a formula in the language of arithmetic, and $\bar{a} \in \mathbb{M}$. Show that if $\mathcal{M} \models \phi(n,\bar{a})$ for all $n < \omega$, then there is an infinite $c \in \mathbb{M}$ s.t. $\mathcal{M} \models \phi(c,\bar{a})$.

My Proof

Let $\mathcal{N}$ be the standard model of Peano arithmetic, and $\pmb{PA}$ the theory of Peano arithmetic so that $\mathcal{N} \models \pmb{PA}$; additionally, we know by definition $\mathcal{M} \models \pmb{PA}$. Suppose for the sake of contradiction that there exists a nonstandard model $\mathcal{M}$ so that $\mathcal{M} \models \phi(n,\bar{a})$ but $\mathcal{M} \not\models \phi(c,\bar{a})$. Since $\mathcal{M} \models \pmb{PA}$, $\mathcal{M} \models \phi(0,\bar{a})$ and by the assumption that $\mathcal{M} \models \phi(n,\bar{a})$ for all $n < \omega$, we know $\mathcal{M} \models \forall x (\phi(x,\bar{a}) \Rightarrow \phi(x + 1, \bar{a}))$. Hence by the induction axiom of $\pmb{PA}$:

$$Ind(\phi) := (\phi(0) \wedge \forall x(\phi(x) \Rightarrow (x+1))) \Rightarrow \forall x \phi(x),$$

we know $\mathcal{M} \models \forall x (\phi(x,\bar{v}))$, therefore $\mathcal{M} \models \phi(c,\bar{a})$. But this contradicts our hypothesis and we conclude: $$\mathcal{M} \models \phi(c,\bar{a}).$$

My Problem

My primary concern is whether I can use the induction principle of Peano arithmetic to argue that I can "reach" this $c$ by induction. Since as I understand, this $c$ lies beyond $\omega$.

Additionally, is there another way to prove the proposition without invoking the details of Peano Arithmetic?

• I have a question: why $\mathcal{M}\models \phi(n,\overline{a})$ for all $n<\omega$ implies $\mathcal{M}\models \forall x (\phi(x,\overline{a})\to \phi(x+1,\overline{a}))$? Feb 13, 2015 at 4:25
• I presume it's because all the n's less than $\omega$ can be reached by induction, ie: $0,1,\ldots$. So for every $x$, you can always count up one and $\phi$ is still satisfied. Feb 13, 2015 at 4:30
• I can't find a reason that $\phi(x,\overline{a})\to\phi(x+1,\overline{a})$ holds for nonstandard $x$. Feb 13, 2015 at 4:35
• So I presume the reason is by the induction principle which is an axiom of PA. This is the part where I am iffy about. Since the principle says if it's true for 0 and x, x+1, then it's true for all x. I presume the universal quantifier is over all elements of $\mathbb{M}$, which includes c. Feb 13, 2015 at 4:47
• Ah, I realize it. If $\phi(x,a)$ does not hold for all nonstandard $x$, then $\lnot\phi(x+1,a)\to \lnot\phi(x,a)$ holds for all nonstandard $x$ so you can guarantee that $\forall x[ \phi(x,a)\to\phi(x+1,a)]$ holds on $\mathcal{M}$. Feb 13, 2015 at 4:50

To me, Rick's proof is correct, but does not give enough details. Here is a complete proof. Let's assume we have a formula $$\psi$$ such that : $$\forall n < \omega, \mathcal{M} \vDash \psi(n)$$ Let's show that there exists a non standard integer satisfying $$\psi$$. By contradiction, we assume that all non standard integers do not satisfy $$\psi$$.

1. We have $$\mathcal{M} \vDash \psi (0)$$. For $$n < \omega$$, we have $$\mathcal{M} \vDash \psi(n+1)$$, so we also have $$\mathcal{M} \vDash \psi(n) \rightarrow \psi(n+1)$$
2. By hypothesis, for a non standard integer $$b$$ we have $$\mathcal{M}\nvDash \psi(b)$$, so, exfalso, we have $$\mathcal{M} \vDash \psi(b) \rightarrow \psi(b+1)$$ (the left side of implication is false).
3. By induction hypothesis, that is true in $$\mathcal{M}$$ as it is a model of Peano's arithmetic, we have $$\forall x, \mathcal{M} \vDash \psi(x)$$, which contradicts our hypothesis. So there exists at least one non standard integer satisfying $$\psi$$.

One can easily extend this proof to show that once we have $$\mathcal{M} \vDash \psi(a)$$ for some-non standard integer $$a$$, then we have $$\forall b < a, \mathcal{M} \vDash \psi(b)$$.

Have a good day !

The induction principle is OK in $\mathcal M$, but its hypothesis is not necessarily satisfied in your situation. You inferred from (1) $\mathcal M\models\phi(n,\bar a)$ for all $n<\omega$ that (2) $\mathcal M\models\forall x\,(\phi(x,\bar a)\implies\phi(x+1,\bar a))$. Unfortunately, (1) does not imply (2). The error arises because the variable $x$ in (2) ranges over the whole universe of $\mathcal M$, not just over numbers $n<\omega$.

Note also that, if your argument were correct, it would give that $\mathcal M\models\phi(c,\bar a)$ for all elements $c$ of $\mathcal M$. That conclusion is not in general right; the correct conclusion is that some infinite $c$ in $\mathcal M$ satisfies $\mathcal M\models\phi(c,\bar a)$. (One can do a bit better; there is some infinite $d$ in $\mathcal M$ such that $\mathcal M\models\phi(c,\bar a)$ for all $c\leq d$.)

• I think that if we assume (for an argument by contradiction), that $\mathcal{M} \vDash \phi(n,a)$ for all $n \in \mathbb{N}$ but $\mathcal{M} \nvDash \phi(c,a)$ for all $c \in \mathcal{M}\setminus\mathbb{N}$, then the argument of the OP will work. (In the assumption, I have added the quantification for the $n$ and the $c$ which were missing in the original question.) To put it otherwise: If we assume $\mathcal{M}\vDash \phi(n,a)$ only for standard numbers $n$, then (1) would imply (2). Or am I wrong? Feb 13, 2015 at 16:10
• @russoo Yes. From the assumptions that $\phi(x,\bar a)$ holds in $\mathcal M$ for all standard $x$ and that it holds for no nonstandard $x$, then we can infer the assertion (2) in my answer. The OP may or may not have had the second part of the assumption in mind earlier in the argument (as you noticed, the quantification of $c$ is missing), but by the time he wanted (2) he was claiming that it followed from just (1). Feb 13, 2015 at 16:19

Summing up the answer from Andreas and the comments above, here is a full answer to the question.

Assume for contradiction that $$\mathscr M \not\models \psi(c,\overline a)$$ for all non-standard $$c \in |\mathscr M|$$. By assumption, we have that $$\mathscr M \models \psi(0, \overline a)$$ and $$\mathscr M \models (\psi(n, \overline a) \rightarrow \psi(n+1, \overline a))$$, so by the induction axiom of PA it follows that $$\mathscr M \models \forall x (\psi(x, \overline a))$$. In particular, $$\mathscr M \models \psi(c, \overline a)$$ for all non-standard $$c \in |\mathscr M|$$, a contradiction.

Below is a more "positive" version of the Overspill Lemma, which does not require a proof by contradiction which slightly eases the proof.

A $$\mathsf{PA}$$ model $$\mathcal{M}$$ is isomorphic to $$\mathbb{N}$$ if and only if there is a formula $$\varphi(x)$$ such that $$e < \omega \iff \mathcal{M} \vDash \varphi(e)$$ for every $$e \in \mathcal{M}$$.

Proof: $$(\Longrightarrow)$$ If $$\mathcal{M} \cong \mathbb{N}$$ then there is indeed such a formula $$\varphi(x)$$, namely $$x = x$$.

$$(\Longleftarrow)$$ Assume there is a formula $$\varphi(x)$$ with $$e < \omega \iff \mathcal{M} \vDash \varphi(e)$$, then note that we have:

• $$0 < \omega$$ and therefore $$\mathcal{M} \vDash \varphi(0)$$.
• Obviously we have $$e < \omega \Rightarrow e + 1 < \omega$$, which is equivalent to $$\mathcal{M} \vDash \varphi(e) \Rightarrow \mathcal{M} \vDash \varphi(e+1)$$ by our assumption. Overall this shows $$\mathcal{M} \vDash \forall x \; (\, \varphi(x) \to \varphi(x+1) \,)$$.

Together with the induction principle the above bullet points imply the truth of $$\mathcal{M} \vDash \forall x \; \varphi(x)$$, further giving us $$\mathcal{M} \vDash \forall x \; \varphi(x) \iff \forall e \in \mathcal{M} : \mathcal{M} \vDash \varphi(e) \iff \forall e \in \mathcal{M} : e < \omega$$ showing that all of the elements in $$\mathcal{M}$$ are standard numbers, i.e. $$\mathcal{M} \cong \mathbb{N}$$. $$\hspace{1em} \Box$$

• Institutionistically, the above proof establishes a stronger version of Overspill.
• The usual Overspill Lemma is implied by the contrapostion of the above constructive result.
• $$e < \omega$$ should be understood as $$\exists \, n \in \mathbb{N} \; (e = S^n 0)$$.

This is my current proof. It incorporates suggestions from comments above and elsewhere.

First recall that $\phi(x) \rightarrow \phi(x+1)$ is short hand for $\neg \phi(x) \vee \phi(x+1)$. Now suppose for some infinite $c$, we have $\mathcal{M} \models \neg\psi(c,\bar{a})$, then the statement:

$$\mathcal{M} \models \neg\psi(c,\bar{a}) \vee \mathcal{M} \models (c + 1, \bar{a}),$$

is true since the first statement of the disjunction is true by definition. But by definition:

$$\mathcal{M} \models \neg\psi(c,\bar{a}) \vee \mathcal{M} \models (c + 1, \bar{a}) \Rightarrow \mathcal{M} \models \psi(c,\bar{a}) \rightarrow \psi(c + 1, \bar{a}),$$

hence the implication is true for all infinite $c$. On the other hand, we know $\mathcal{M} \models \psi(n,\bar{a})$ for all $n < \omega$ so $\mathcal{M} \models \psi(n+1,\bar{a})$. Hence we have:

$$\mathcal{M} \models \neg\psi(n,\bar{a}) \vee \mathcal{M} \models (n + 1, \bar{a}) \Rightarrow \mathcal{M} \models \psi(n,\bar{a}) \rightarrow \psi(n + 1, \bar{a}),$$

where the term on the left hand side of of $\Rightarrow$ is true because the term to the right of the disjunction is true by assumption. But we know all terms in $\mathbb{M}$ is either of form $n < \omega$ or some infinite $c$, so we can say:

$$\forall x (\psi(x,\bar{a}) \rightarrow \psi(x+1,\bar{a})) \rightarrow \forall x \psi(x,\bar{a}).$$ Hence $\mathcal{M} \models \psi(c,\bar{a})$ for some infinite $c \in \mathbb{M}$.

• For anyone seeing this: this solution given by chibro2 here is not right! On the one hand, the assumption that there is some infinite $c$ such that $\mathscr M \models \neg\psi(c, \overline a)$ plays absolutely no role and is introduced for no reason. On the other hand the middle implications are actually "iffs" as they are a re-statement of the definition of $\rightarrow$; moreover, the last implication is clearly false under the assumption that there exists $c$ such that $\mathscr M \models \neg\psi(c, \overline a)$.
– Rick
Jun 1, 2020 at 11:17