# Does Zorn Lemma imply the existence of a (not unique) maximal prolongation of any solution of an ode?

Let be given a map $F:(x,y)\in\mathbb{R}\times\mathbb{R}^n\to F(x,y)\in\mathbb{R}^n$.
Let us denote by $\mathcal{P}$ the set whose elements are the solutions of the ode $y'=F(x,y)$, i.e. the differentiable maps $u:J\to\mathbb{R}^n$, where $J\$ is some open interval in $\mathbb{R}\$, s.t. $u'(t)=F(t,u(t))$ for all $t\in J$.
Let $\mathcal{P}$ be endowed with the ordering by extension.

In order to prove that any element of $\mathcal{P}$ is extendable to a (not unique) maximal element, without particular hypothesis on $F$, I was wondering if the Zorn lemma can be used.

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I vaguely recall there might be something in Coddington & Levinson about this, in a section about existence theorems. –  Michael Hardy Oct 18 '11 at 21:15
You'll probably want to restrict your attention explicitly to functions defined on connected subsets of $\mathbb R$. Otherwise if you start with, say, a solution defined on the open unit interval, the "maximal extension" you get could be the original solution extended periodically to $\mathbb R\setminus\mathbb N$ with jump discontinuities at each integer. And that isn't very interesting... –  Henning Makholm Oct 18 '11 at 21:38
Dear Henning Makholm, I have edited the question after your comment. –  Giuseppe Tortorella Oct 18 '11 at 21:49
You need the existence theorem itself to guarantee $\mathcal{P}$ is not empty. The next issue, to apply Zorn's lemma you need upper bounds for each chain -- showing this strikes me as a big problem. –  Bill Cook Oct 18 '11 at 23:19
@Bill, existence of a partial solution is assumed -- it is implicit in stating the proposition as "any element of $P$ ...". As for upper bounds, what could go wrong? Under the common definition of a function as the set of pairs $\langle x,f(x)\rangle$, the union of all functions in a chain will -- as far as I can see -- be a function with the required property. –  Henning Makholm Oct 20 '11 at 19:28

Yes, Zorn's Lemma should be all you need. Take the set of partial solutions that extend your initial solution, and order them by the subset relation under the common definition of a function as the set of pairs $\langle x, f(x)\rangle$. Then the union of all functions in a chain will be another partial solution, so Zorn's Lemma applies.