# How to find this upper bound to apply Zorn lemma

I want to prove that every solution of $x'=f(t,x)$ can be enlarged in a maximal solution.

In order to prove that , I use the Zorn lemma defining $X=\{(\varphi,I);\varphi:I\to \mathbb R^n$ is solution of $x'=f(t,x);x(t_0)=x_0\}$. Now let's define a partial order $X$: $(\varphi_1,I_1)\leq(\varphi_2,I_2)\leftrightarrow I_1\subset I_2$ and $\varphi_2|I_1=\varphi_1$, if I take a chain in this set, I couldn't find the upper bound of this chain.

I need help

Thanks

• Did you mean to say that $\varphi_2|I_1=\varphi$? – Asaf Karagila Jul 26 '13 at 9:47
• Please specify what kind of sets the domains $I$ are. Also, I would be curious to see why Zorn's lemma is needed for this. – Bob Terrell Jul 26 '13 at 12:25
• @AsafKaragila I will edit it – user42912 Jul 27 '13 at 6:07
• @BobTerrell $I_1$ and $I_2$ are intervals – user42912 Jul 27 '13 at 6:10
• I think you have to say that the $I$ are open intervals containing $t_0$. – Bob Terrell Jul 27 '13 at 12:37

Suppose $\{(\varphi_{\alpha}, I_{\alpha}) \ | \ \alpha \in A \}$ is a chain. Take the function $\varphi$ which has domain $\bigcup_{\alpha \in A}I_{\alpha}$ and $\varphi(x) = \varphi_{\alpha}(x)$ for any $\alpha$ such that $x \in I_\alpha$ . Now show this is an upper bound and well defined.
If you think about functions are sets of ordered pairs, then the requirement that $\varphi_2|I_1=\varphi$ is essentially saying $\varphi\subseteq\varphi_2$.