Show $\inf\{t\ge0:x(t-)\in B\text{ or }x(t)\in B\}\le T$ iff $x(T)\in B=\bigcap_nB_n$ or $\forall n:\exists s\in[0,T):x(s)\in B_n$

Question

Let $(E,d)$ be a metric space, $x:[0,\infty)\to E$ be càdlàg (right-continuous with left-limits), $x(0-):=x(0)$, $x(t-):=\lim_{s\to t-}x(s)$ for $t>0$, $B\subseteq E$ be nonempty and $$\tau:=\inf\underbrace{\{t\ge0:x(t-)\in\overline B\text{ or }x(t)\in\overline B\}}_{=:\:I}.$$

Note that $$B_n:=\left\{x\in E:d(x,B)<\frac1n\right\}$$ is open for all $n\in\mathbb N$ and $(B_n)_{n\in\mathbb N}$ is nondecreasing with $$\bigcap_{n\in\mathbb N}B_n=\overline B\tag1.$$

Let $t\ge0$. How can we show that

1. $\tau\le t$;
2. $x(t)\in\overline B$ or for all $n\in\mathbb N$, there is a $s\in\mathbb Q\cap[0,t)$ with $x(s)\in B_n$

are equivalent?

I really struggle to show even that (1.) implies (2.). What's the easiest way to show this?

$\tau\le t$ clearly implies that $I\cap[0,t+\delta)\ne\emptyset$ for all $\delta>0$. But is this useful?

Or maybe we need to assume $x(t)\not\in\overline B$ which yields $$x(t)\in E\setminus\overline B=\bigcup_{n\in\mathbb N}(E\setminus B_n)\tag2;$$ but again, I don't know how this could be of use ...

Answer 1

Assume (2.) is satisfied. If $x(t)\in\overline{B}$ we are done. If not we may pick a sequence $(s_n)$ in $\mathbb{Q}\cap[0,t)$ with $x(s_n)\in B_n$ for all $n\in\mathbb{N}$. From here we choose a monotone subsequence $(s_{n_k})$ and note that $s=\lim_{k\to\infty}s_{n_k}$ exists because the sequence is bounded. Moreover, $y=\lim_{k\to\infty}x(s_{n_k})$ exists since $x$ has right and left limits at all points. Then $d(y,B)=\lim_{k\to\infty}d(x(s_{n_k}),B)\leq\lim_{k\to\infty}1/n_k=0$, so $y\in\overline{B}$. Since we have either $y=x(s-)$ or $y=x(s)$ we find that $s\in I$. Hence, $\tau\leq s\leq t$.

Assume now that (2.) is not satisfied. This means that $x(t)\notin\overline{B}$ and that there exists $n\in\mathbb{N}$ such that $x(s)\notin B_n$ for all $s\in\mathbb{Q}\cap[0,t)$. For any $u\in[0,t)$ we can write $u=\lim_{k\to\infty}s_k$ where $(s_k)$ is a non-increasing sequence in $\mathbb{Q}\cap[0,t)$. By right-continuity we have $x(u)=\lim_{k\to\infty}x(s_k)\in B_n^\mathsf{c}$ since $B_n^\mathsf{c}$ is closed. Hence we have $x(u)\notin\overline{B}$ for all $u\in[0,t]$. Similarly, for any $u\in[0,t]$ we may approximate with a non-decreasing sequence and obtain $x(u-)\in B_n^\mathsf{c}$ which shows that $x(u-)\notin\overline{B}$. We have thus shown that $\tau\geq t$. Suppose now that $\tau=t$. Then we may pick a non-increasing sequence $(s_n)$ in $(t,\infty)$ with $\lim_{n\to\infty}s_n=t$ such that for any $n\in\mathbb{N}$ we have $x(s_n)\in\overline{B}$ or $x(s_n-)\in\overline{B}$. We construct a new sequence $(s_n')$ in $(t,\infty)$ in the following way: If $x(s_n)\in\overline{B}$ let $s_n'=s_n$. If $x(s_n)\notin\overline{B}$ we have $x(s_n-)\in\overline{B}$ so we may pick $s_n'\in(t,s_n)$ such that $x(s_n')\in B_n$. We now have $s_n'>t$ for all $n\in\mathbb{N}$ and $s_n'\to t$, and thus $x(s_n')\to x(t)$ by right-continuity. Also, for each $n\in\mathbb{N}$ we have $x(s_n')\in B_n$ (recall $\overline{B}\subseteq B_n$) so $d(x(t),B)=\lim_{n\to\infty}d(x(s_n'),B_n)\leq\lim_{n\to\infty}1/n=0$, showing that $x(t)\in\overline{B}$ which is a contradiction. We conclude that $\tau>t$.