[proofplan] We prove the contrapositive. If a maximal solution remains in a fixed compact subset $C\subset U$ arbitrarily close to its finite right endpoint $b$, then continuity of $F$ gives a uniform bound on the vector field along the relevant compact time-state cylinder. The integral formulation of the ODE then makes $x$ Lipschitz on a left-neighbourhood of $b$, so $x(t)$ has a limit $y\in C$ as $t\to b^-$. A local solution starting from $(b,y)$ can then be concatenated with $x$, producing a proper extension beyond $b$ and contradicting maximality. [/proofplan]

[step:Assume the solution remains in one compact set before $b$] Suppose, toward a contradiction, that there exist a compact set $C\subset U$ and a number $a<b$ such that \begin{align*} x(t)\in C \end{align*} for every $t\in J$ satisfying $a<t<b$. Since $b=\sup J$ and $a<b$, there exists $t_0\in J$ with $a<t_0<b$. Since $b$ is not the right endpoint of $I$, there exists $s\in I$ with $b<s$. The set $J$ is contained in $I$, so $t_0\in I$. Because $I$ is an interval and $t_0<s$, we have \begin{align*} [t_0,s]\subset I. \end{align*} In particular, $b\in I$ and $[t_0,b]\subset I$. [/step]

[step:Bound the vector field on the compact time-state cylinder]Define \begin{align*} K:=[t_0,b]\times C\subset I\times U. \end{align*} The interval $[t_0,b]$ is compact in $\mathbb{R}$, and $C$ is compact in $\mathbb{R}^n$, hence $K$ is compact in $\mathbb{R}\times\mathbb{R}^n$. Since $F:I\times U\to\mathbb{R}^n$ is continuous, its restriction to $K$ is bounded. Thus there exists a constant $M\ge 0$ such that \begin{align*} |F(t,z)|\le M \end{align*} for every $(t,z)\in K$.[/step]

[guided]The contradiction hypothesis says that, once time is larger than $a$ and still less than $b$, the solution never leaves the compact state set $C$. We choose $t_0\in J$ with $a<t_0<b$ so that all later times $t\in J$ with $t_0\le t<b$ satisfy $x(t)\in C$. The time interval relevant to the argument is $[t_0,b]$. This interval lies inside the ambient interval $I$: indeed, $t_0\in J\subset I$, and by hypothesis there is $s\in I$ with $b<s$; since $I$ is an interval, every point between $t_0$ and $s$ lies in $I$. Therefore \begin{align*} [t_0,b]\subset I. \end{align*} Now define the compact time-state cylinder \begin{align*} K:=[t_0,b]\times C. \end{align*} The product of compact subsets of Euclidean spaces is compact, so $K$ is compact in $\mathbb{R}^{n+1}$. Also $K\subset I\times U$, because $[t_0,b]\subset I$ and $C\subset U$. Since $F:I\times U\to\mathbb{R}^n$ is continuous, the extreme value theorem gives a finite bound on $|F|$ over $K$. Thus there exists $M\ge 0$ such that \begin{align*} |F(t,z)|\le M \end{align*} for every $(t,z)\in K$. This bound is the quantitative input needed to prove that the curve cannot oscillate infinitely fast as $t$ approaches $b$ from the left.[/guided]

[step:Use the integral equation to obtain a left limit at $b$] Let $r,u\in J$ satisfy $t_0\le r<u<b$. By the integral form of a solution, applied on the compact interval $[r,u]\subset J$, we have \begin{align*} x(u)-x(r)=\int_r^u F(\tau,x(\tau))\,d\mathcal{L}^1(\tau). \end{align*} For every $\tau\in[r,u]$, the contradiction hypothesis gives $x(\tau)\in C$, and $\tau\in[t_0,b]$. Hence $(\tau,x(\tau))\in K$, and therefore \begin{align*} |x(u)-x(r)|\le \int_r^u |F(\tau,x(\tau))|\,d\mathcal{L}^1(\tau)\le M(u-r). \end{align*} It follows that $x$ is uniformly Lipschitz on $J\cap[t_0,b)$. Hence, if $(t_k)_{k\in\mathbb{N}}$ is any sequence in $J\cap[t_0,b)$ with $t_k\to b$, then $(x(t_k))_{k\in\mathbb{N}}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $\mathbb{R}^n$. Since $\mathbb{R}^n$ is complete, it has a limit. The Lipschitz estimate also shows that this limit is independent of the sequence. Denote the resulting left limit by $y\in\mathbb{R}^n$: \begin{align*} \lim_{\substack{t\to b^-, t\in J}} x(t)=y. \end{align*} Because $x(t)\in C$ for all $t\in J$ with $t_0\le t<b$ and $C$ is closed as a compact subset of $\mathbb{R}^n$, we have $y\in C\subset U$. [/step]

[step:Start a local solution from the limiting state at time $b$] Since $y\in U$ and $U\subset\mathbb{R}^n$ is open, choose $\rho>0$ such that $\overline{B}(y,\rho)\subset U$. Since there exists $s\in I$ with $b<s$ and $I$ is an interval, choose $\delta>0$ such that $b+\delta<s$; then $[b,b+\delta]\subset I$. On the time-state rectangle $[b,b+\delta]\times \overline{B}(y,\rho)$, the restriction of $F$ is continuous and locally Lipschitz in the state variable. The one-sided local existence theorem for ordinary differential equations therefore gives a number $\varepsilon\in(0,\delta]$ and a solution \begin{align*} z:[b,b+\varepsilon]\to U \end{align*} of \begin{align*} \dot{z}(t)=F(t,z(t)),\qquad z(b)=y. \end{align*} Here the differential equation at the left endpoint $b$ is interpreted by the right derivative, matching the solution convention in the theorem statement. [/step]

[step:Concatenate the old and new solutions to contradict maximality] Define an interval \begin{align*} \widetilde{J}:=J\cup[b,b+\varepsilon]. \end{align*} We verify that this union has no gap. If $q\in J$ and $q<r<b$, then $r\in J$ because $J$ is an interval and there exists $p\in J$ with $r<p<b$, using $b=\sup J$. Hence $J$ contains all points of $I$ lying between any sufficiently late point of $J$ and $b$, so adjoining $[b,b+\varepsilon]$ produces an interval. Since $[b,b+\varepsilon]\subset I$, we have $\widetilde{J}\subset I$, and because $b+\varepsilon\in\widetilde{J}$ while $b+\varepsilon\notin J$, the containment $J\subsetneq\widetilde{J}$ is proper. Define \begin{align*} \widetilde{x}:\widetilde{J}\to U \end{align*} by \begin{align*} \widetilde{x}(t)=x(t)\quad\text{for }t\in J,\ t<b, \end{align*} and \begin{align*} \widetilde{x}(t)=z(t)\quad\text{for }t\in[b,b+\varepsilon]. \end{align*} If $b\in J$, then the previously established limit gives $x(b)=y=z(b)$, because $x$ is continuous. If $b\notin J$, the same equality gives the continuous value inserted at the joining time. Thus $\widetilde{x}$ is continuous at $b$ and agrees with a solution on each side of $b$. For every compact subinterval of $\widetilde{J}$ not crossing $b$, the differential equation holds because either $\widetilde{x}=x$ or $\widetilde{x}=z$. At $b$, the right derivative of $\widetilde{x}$ is the right derivative of $z$, equal to $F(b,y)$, and the left derivative, when required by the endpoint convention for solutions, follows from the left-hand equation for $x$ together with continuity of $F$ and $x(t)\to y$. Hence $\widetilde{x}$ is a solution on $\widetilde{J}$. This solution extends $x$ to a strictly larger interval contained in $I$, contradicting maximality of $x:J\to U$. Therefore the assumed compact set $C$ and number $a<b$ cannot exist. Equivalently, for every compact set $C\subset U$ and every $a<b$, there exists $t\in J$ such that $a<t<b$ and $x(t)\notin C$. [/step]