A non-extendable solution leaves every compact subset of the domain near its finite endpoint; an autonomous solution trapped in a compact subset of its state domain cannot break down there

Let $n\ge 1$, let $D\subseteq \mathbb{R}\times {\mathbb{R}}^n$ be open, and let $F:D\to {\mathbb{R}}^n$ be continuous and locally Lipschitz in the state variable in the sense of Local Lipschitz continuity in the state variable, locally uniform in time and parameters. Let $t_0<b$ be reals, let $x_0\in {\mathbb{R}}^n$ with $(t_0,x_0)\in D$, and let $x:[t_0,b)\to {\mathbb{R}}^n$ be a solution of the initial value problem $x^{\mathrm{\prime }}=F(t,x)$, $x(t_0)=x_0$ (First-order systems, initial value problems, and solutions on intervals). Suppose $x$ admits no extension past $b$: there do not exist $\epsilon >0$ and a solution $\tilde{x}:[t_0,b+\epsilon )\to {\mathbb{R}}^n$ of the same initial value problem with $\tilde{x}(t)=x(t)$ for every $t\in [t_0,b)$. Then:

1. for every compact $K\subseteq D$ and every $\eta >0$ there is $t\in (b-\eta ,b)\cap [t_0,b)$ with $(t,x(t))\mathrm{\notin }K$; that is, the graph of $x$ leaves every compact subset of $D$ that could contain it arbitrarily close to $b$;
2. if $D=\mathbb{R}\times U$ for an open $U\subseteq {\mathbb{R}}^n$, $F$ is autonomous, $F(t,z)=G(z)$ with $G:U\to {\mathbb{R}}^n$ continuous and locally Lipschitz in the state variable, and there are a compact $K_0\subseteq U$ and $\delta >0$ with $x(t)\in K_0$ for all $t\in (b-\delta ,b)\cap [t_0,b)$, then the no-extension hypothesis is false: $x$ does extend past $b$.