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By global Peano existence theorem I mean the existence of a maximal interval of solution of a first order ODE $x'=f(x,t)$ with continuous $f$. The proofs of the global Peano Theorem found in the literature often simply appeal to Zorn’s Lemma; eg. Theorem 4.7 in Ganesh

S. S. Ganesh, Lecture Notes on Ordinary Differential Equations, Annual Foundation School IIT Kanpur, December 3 - 28, 2007, 34 pp. https://www.math.iitb.ac.in/~siva/afs07.pdf

The more careful proofs depend on ADC, usually without mentioning it explicitly. Hale

J. Hale, Ordinary Differential Equations, 2nd Edition, R. E. Krieger Publ. Co., Florida, 1980

 in his proof of global Peano Theorem (Theorem 2.1, p. 17) writes:

“...there is a monotone increasing sequence {bn} constructed as above so that the solution $x(t)$ of (1.1) on $[a, b]$ has an extension to the interval $[a, b_n]$ and $(b_n, x(b_n))$ is not in $V_n$. Since the $b_n$ are bounded above, let $\omega = \lim_{n\to\infty} b_n$. It is clear that $x$ has been extended to the interval $[a, \omega)$...” 

What is actually clear is that his construction yields solutions $x_n(t)$ on $[a, b_n]$ for each $n$, and each $x_n(t)$ has extensions to some $x_{n+1}(t)$.  ADC is needed to justify the existence of $x(t)$.  Similarly Hartman

P. Hartman, Ordinary Differential Equations, 2nd Edition, SIAM, Philadelphia, 2002

in the proof of II, 3.1, p. 13, constructs an increasing sequence $\{b_n\}$ such that any solution on $[a, b_n]$ has an extension to a solution on $[a, b_{n+1}]$.  ADC is needed to justify the existence of a solution on $[a, \omega^+]$ for $\omega^+ = \lim_{n\to\infty} b_n$.  In the proof of III, Lemma 3.1, a key step to the proof of III, 3.1 (Osgood’s Theorem), ACC is used implicitly to choose the sequence $\{u_n(t)\}$.  Similar unacknowledged use of ADC appears on pp. 355–356 in

J. Kurzweil, Ordinary Differential Equations. Introduction to the Theory of Ordinary Differential Equations in the Real Domain, translated from the Czech by M. Basch, Studies in Applied Mechanics, 13, Elsevier Scientific Publishing, Amsterdam, 1986.

Is there a published proof of global Peano existence that's valid in ZF?

Note. Our article can be found at Hrbacek, K.; Katz, M. "Peano and Osgood theorems via effective infinitesimals." Journal of Logic and Analysis 15 (2023). https://arxiv.org/abs/2311.01374

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    $\begingroup$ I don't know if it's written out explicitly somewhere, but it should be provable in ZF by a fairly direct absoluteness argument. The vague idea is that you can form a real number $\alpha$ that codes the data associated to the specific IVP. Then in $L[\alpha]$, choice holds and you can solve the IVP there. The solution will then have a unique extension to a differentiable function in $V$ which will still solve the problem. $\endgroup$ Commented Oct 4, 2023 at 12:45
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    $\begingroup$ I'd be happy to write it out as an answer, although I don't have time at this moment. I actually wrote up an explanation of almost this specific case in an answer about a month ago. Unfortunately though, a more precise explanation is probably the topic of an expository paper (which I have been contemplating). $\endgroup$ Commented Oct 4, 2023 at 12:57
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    $\begingroup$ Yes it would, but as I indirectly discuss in that answer Peano's theorem (including the maximality condition) should be $\Pi^1_4$. $\endgroup$ Commented Oct 4, 2023 at 13:06
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    $\begingroup$ I should mention that we have a proof of global Peano in SPOT that's currently submitted. $\endgroup$ Commented Oct 4, 2023 at 13:06
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    $\begingroup$ Stephen G. Simpson has shown that over RCA_0, Peano's existence theorem is equivalent to WKL. I would imagine that adding the maximality condition is possible without too much effort $\endgroup$
    – Holo
    Commented Oct 4, 2023 at 14:05

1 Answer 1

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First of all, as Holo mentioned in the comments, it's already known that the pure existential statement of the Peano existence theorem can be shown in $\mathsf{WKL}_0$, which is significantly weaker than $\mathsf{ZF}$.

Also, as I mentioned, this answer is going to be more or less a rehash of the ideas I discussed in my answer here, but I'll tweak the presentation somewhat to try to get at the intuition behind deciding whether Shoenfield absoluteness applies to a given statement in analysis.


I'm going to use the version of the theorem given on Wikipedia with an extra stipulation that we want the resulting solution to be maximal. In other words, I'm going to sketch an argument using Shoenfield absoluteness that the following statement is provable in $\mathsf{ZF}$.

(Peano existence theorem) For any open $D \subseteq \mathbb{R}^2$, continuous $f:D \to \mathbb{R}$ and initial condition $\langle t_0,x_0\rangle \in D$, there is an open interval $I \subseteq \mathbb{R}$ with $t_0 \in I$ and a differentiable function $X: I \to \mathbb{R}$ such that $X(t_0)=x_0$ and $X'(t) = f(X(t),t)$ for all $t \in I$, and no strictly larger $I_\ast \supset I$ has such a function extending $f$.

So Shoenfield absoluteness allows us automatically translate $\mathsf{ZFC}$ proofs of low quantifier complexity statements about countably coded objects to $\mathsf{ZF}$ proofs. The specific form I'm going to use is this:

If $\varphi$ is a theorem of $\mathsf{ZFC}$ that can be written in the form $$\forall x(P(x) \to \exists y(Q(x,y) \wedge \forall z(R(x,y,z) \to \exists w S(x,y,z,w))))$$ where $x$, $y$, $z$, and $w$ range over some collections of countably coded objects and $P$, $Q$, $R$, and $S$ are Borel conditions, then $\varphi$ is a theorem of $\mathsf{ZF}$.

Sentences of this form are called $\Pi^1_4$, where $\Pi$ corresponds to the fact that the leading quantifier is $\forall$, $1$ corresponds to the fact that we are quantifying over real numbers (countably coded objects), and $4$ refers to the fact that we have $4$ alternating quantifiers.

What I find notable about this is that it doesn't matter if the original proof uses things that become much more difficult to formalize without choice (like measure theory) or that explicitly require some choice (like some applications of compactness). We can systematically convert these to $\mathsf{ZF}$ proofs (although these proofs, once unwound, may not be that pleasant).

Rather than give a fully general definition of what I mean by 'countably coded objects,' I will list some specific examples that are relevant to the Peano existence theorem.

  • A real number is a countably coded object.
  • An open subset of $\mathbb{R}^n$ is a countably coded object (expressed as a union of a sequence of open balls with rational centers and rational radii).
  • A continuous function to $\mathbb{R}$ with open domain is a countably coded object (expressed by its domain and a sequence of piecewise affine functions whose graphs can be written as finite simplical complexes with rational vertices).
  • Any finite or countably infinite sequence of countably coded objects is a countably coded object.

In the continuous functions the sequence needs to be converging to the continuous function uniformly on compact sets. To make this explicit, we could say that if the sequence is $(f_i)_{i \in \mathbb{N}}$ and its domain is coded as a union of open balls $\bigcup_{j\in \mathbb{N}}B_{<r_j}(x_j)$, then we require that for each $n\in \mathbb{N}$, $i,i' > n$, and $j \leq n$, $\sup_{x \in B_{\leq(1-2^{-n})r_j}(x_j)}|f_i(x)-f_{i'}(x)| \leq 2^{-n}$. (There are many ways we could do this. Also, we do technically need to show in $\mathsf{ZF}$ that any continuous function on an open set can be written as such a sequence, but this is not so hard to show directly.)

The other concept we need to describe Shoenfield absoluteness is that of a 'Borel condition.' For objects naturally living in a Polish space, this actually does correspond precisely to being in a Borel set (justifying the name), but I'm just going to give a list of relevant Borel conditions without defining the notion precisely.

  • Given a countably coded open set $U \subseteq \mathbb{R}^n$ and $x \in \mathbb{R}$, the condition $x \in U$ is Borel.
  • Given countably coded open sets $U$ and $U_\ast$ in $\mathbb{R}^n$, the condition $U \subseteq U_\ast$ is Borel.
  • For countably coded open $U \subseteq \mathbb{R}$, the condition "$U$ is an interval" is Borel.
  • For a countably coded continuous function $f: U \to \mathbb{R}$, $x \in U$, and $r \in \mathbb{R}$, $f(x) = r$ is a Borel condition.
  • For a countably coded continuous function $f: U \to \mathbb{R}$ (with $U\subseteq\mathbb{R}$) and $x$ and $r$ in $\mathbb{R}$, "$f'(x)$ exists and $f'(x) = r$" is a Borel condition.
  • Any Boolean combination of Borel conditions is a Borel condition.
  • Any 'explicit' countable conjunction of Borel conditions is a Borel condition.

If I stated a few more (regarding the ability to compose countably coded continuous functions), we'd be able to prove the next one, but I'll just give it explicitly instead.

  • For any countably coded continuous functions $X: I \to \mathbb{R}$, $Y: I \to \mathbb{R}$, and $f: D \to \mathbb{R}$ (with $I \subseteq \mathbb{R}$ and $D \subseteq \mathbb{R}^2$) and any $t \in I$, $Y(t) = f(X(t),t)$ is a Borel condition.

Okay so now we can finally show that the Peano existence theorem is $\Pi^1_4$. (For the record, I think that it might be possible to get this down to $\Pi^1_3$, but it doesn't matter for applying Shoenfield absoluteness.) Since this is kind of a lot to write out all at once, I'll break it down in the same way as the schematic representation before.

Lemma 2. The Peano existence theorem is equivalent (suppressing variables in the Borel conditions) to $$\forall \langle D,f,t_0,x_0\rangle(\langle t_0,x_0 \rangle \in D \to \exists \langle I,X,Y\rangle(Q \wedge \forall \langle t,I_\ast,X_\ast,Y_\ast\rangle(R \to \exists t_\ast S))),$$ where

  • $D \subseteq \mathbb{R}^2$ and $I,I_\ast \subseteq \mathbb{R}$ are countably coded open sets,
  • $f: D \to \mathbb{R}$, $X,Y: I \to \mathbb{R}$, and $X_\ast,Y_\ast: I_\ast\to \mathbb{R}$ are countably coded continuous functions, and
  • $t_0$, $x_0$, $t$, and $t_\ast$ are real numbers

and

  • $Q$ says "$I$ is an interval and $t_0 \in I$,"
  • $R$ says "$t \in I$, $I_\ast$ is an interval, and $I_\ast \supseteq I$," and
  • $S$ says "$X'(t) = Y(t)$ and $Y(t) = f(X(t),t)$ and $t_\ast \in I_\ast$ and [$I \subseteq I_\ast$ or [$t_\ast \in I$ and $X_\ast(t_\ast) \neq X(t_\ast)$] or $X'_\ast(t_\ast) \neq Y_\ast(t_\ast)$ or $Y_\ast(t_\ast)\neq f(X_\ast(t_\ast),t_\ast)$]."

(The nasty bit in the square brackets at the end should be interpreted as saying that $X_\ast$ fails to be a proper extension of $X$ satisfying the differential equation, possibly witnessed at $t_\ast$.)

Proof. (This is largely an exercise in juggling quantifiers.) First assume that the Peano existence theorem is true. Then we have that for any open $D$, continuous $f:D \to \mathbb{R}$, and $\langle t_0,x_0 \rangle \in D$, there is a differentiable function $X: I \to \mathbb{R}$ (on some open interval $I \ni t_0$) such that $X(t_0) = x_0$ and $X'(t) = f(X(t),t)$ for all $t \in I$. Furthermore we have that no strictly larger interval $I_\ast \supset I$ has such a solution extending $f$. Let $Y : I \to \mathbb{R}$ be the derivative of $X$. Note that since $Y(t) = f(X(t),t)$ for all $t \in I$, $Y$ is continuous, so there is a countable coding for $Y$. (Note that while the obvious proof of this seems to need some amount of choice, this is actually provable in $\mathsf{ZF}$.) We need to argue that $\langle I,X,Y \rangle$ is the triple witnessing the rest of the statement in the proposition. Fix a tuple $\langle t,I_\ast,X_\ast,Y_\ast\rangle$ satisfying $R$. We now need to choose $t_\ast$ to make $S$ true. Since $X$ is a solution the the IVP and $t \in I$, we must have that $X'(t) = Y(t) = f(X(t),t)$, so the first two clauses of $S$ are always satisfied. We proceed by cases.

  • If $I \subseteq I_\ast$, then we can let $t_\ast = t_0$ and we're done.
  • If $I_\ast \supset I$ and $Y_\ast$ is not the derivative of $X_\ast$ on $I_\ast$, then there is a $t_\ast \in I_\ast$ such that $X'_\ast(t_\ast) \neq Y(t_\ast)$, so we can choose this and be done.
  • If $I_\ast \supset I$ and $Y_\ast = X'_\ast$ on $I_\ast$, then by the choice of $X$, $X_\ast$ must either fail to extend $X$ or fail to satisfy the differential equation. If $X_\ast$ fails to extend $X$, then there is a $t_\ast \in I$ such that $X(t_\ast) \neq X_\ast(t_\ast)$. So choosing this $t_\ast$, we are done. Otherwise if $X_\ast$ fails to satisfy the differential equation somewhere in its domain, then there is a $t_\ast \in I_\ast$ such that $Y(t_\ast) \neq f(X_\ast(t_\ast),t_\ast)$, so again we are done.

The reverse argument (that the statement in the proposition implies the Peano existence theorem) is roughly the same. I will note that it is fairly straightforward to see that it implies the existence of a solution (by just choosing $X_\ast = X$, $Y_\ast = Y$, and $t_\ast = t_0$ for the last quantifier). Maximality is also fairly straightforward. $\square$

So now finally by Shoenfield absoluteness we get the desired statement

Proposition. $\mathsf{ZF}$ proves the Peano existence theorem (with the maximality condition).

Proof. The statement in Lemma 2 is $\Pi^1_4$, so this follows by Shoenfield absoluteness and the ordinary proof in $\mathsf{ZFC}$. $\square$

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  • $\begingroup$ Can you elaborate on what "the pure existential statement" means? My understanding is that due to issues of coding, the WKL$+)$ version of Peano existence is not really equivalent to the usual ZF version. This is of course true of many results, not merely Peano existence theorem. $\endgroup$ Commented Oct 5, 2023 at 8:24
  • $\begingroup$ This seems like a bit of a miracle. Can you comment briefly why countable additivity of the Lebesgue measure is not $\Pi_4^1$ ? $\endgroup$ Commented Oct 5, 2023 at 8:30
  • $\begingroup$ @MikhailKatz by "pure existential statement" James means Peano's existence theorem without the additional "global" condition (that there is a maximal interval). Not sure what you mean by "equivalent to the usual ZF version", ZF proves everything that ${\sf {WKL}}_0$ can prove when talking about sets of naturals (i.e. reals) $\endgroup$
    – Holo
    Commented Oct 5, 2023 at 9:39
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    $\begingroup$ @MikhailKatz Lebesgue measurable sets are not countably coded $\endgroup$
    – Holo
    Commented Oct 5, 2023 at 9:46
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    $\begingroup$ @MikhailKatz everything in Peano's theorem is countably coded, so you don't lose anything when translating it to ZF. Sometimes you do have cases where WKL_0 (or other weak systems) proves statement A but not prove statement B, but ZF- (or other strong system) prove that A is equivalent to B, fortunately it is not the case here, Simpson proved Peano's theorem in it's original form $\endgroup$
    – Holo
    Commented Oct 5, 2023 at 9:52

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