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My book delivers the theorem in the following form:

$G \subseteq \mathbb{R}\times\mathbb{R}^n$ open and $f:G \rightarrow \mathbb{R}^n$ continuous and suffices local lipschitz-condition. Then the following holds:

1) $\forall\ (t_0,x_0) \in G \ \exists\ \varepsilon\ > 0 $ such that the initial value problem $x'=f(t,x), \ x(t_0)=x_0 $ has exactly one solution $x:[t_0 - \varepsilon, t_0 + \varepsilon] \rightarrow \mathbb{R}^n$

2) If $x_1,x_2: I \rightarrow \mathbb{R}^n$ are two solutions to the IVP with $I$ being an interval with $t_0 \in I$. It follows that $x_1=x_2$

My problem is the proof of 2): Let $V:=\{t\in I\ | x_1(t) = x_2(t)\}\Rightarrow V \neq \emptyset$ since $t_0 \in V$. We want so show that V is clopen to conclude $V=I$. We already know that $V$ is closed. To show openness we consider $t_0 ^\star \in I $. With 1) follows: $\exists\ \varepsilon^\star$>0 such that the IVP $$x'=f(t,x), x(t_0)=x_0^\star:=x_1(t_0^\star)=x_2(t_0^\star) $$ has a unique solution on the interval $[t_0^\star -\varepsilon^\star,t_0^\star +\varepsilon^\star ]$. It follows that $x_1(t)=x_2(t)$ for all$t \in J = I \cap (t_0^\star -\varepsilon^\star,t_0^\star +\varepsilon^\star )$. With that the openness of V is concluded. My question ist, why can I assume that $x_1(t_0^\star) = x_2(t_0^\star)$? Since I only know that for $t_0$.

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    $\begingroup$ you want to show that the set where equality holds is open, so you take some $t$ from that set. By definition of that set the solutions are equal in that point. $\endgroup$ – Thomas Sep 23 '17 at 10:56
  • $\begingroup$ But I only know that about $t_0^\star$ $\endgroup$ – dba Sep 23 '17 at 11:05
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The proof as presented shows directly that $t_0^*$ is an inner point of $I$. As $t_0^*\in I$ is arbitrary, this means that every point of $I$ is an inner point, that is, $I$ is open.

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