How can we characterize $\tau\le t$ for $\tau:=\{t\ge0:x(t)\in B\}$?

Question

Let

• $E$ be a set;
• $x:[0,\infty)\to E$;
• $B\subseteq E$ and $$I:=\{t\ge0:x(t)\in B\};$$
• $\tau:=\inf I$.

Let $t\ge0$. Are are we able to show that

1. $\tau\le t$;
2. $\exists s\in I:s\le t$

are equivalent? If not, can we fix (1.) and/or (2.) in a sensible way?

(2.) clearly implies (1.). Assume (1.) and $$\forall s\in I:s>t\tag3.$$ We can immediatlely conclude that $\tau=t$ and hence $$\tau\not\in I\tag4.$$ So, unless we are assuming $\tau\in I$, this doesn't seem to yield a contradiction.

I'm willing to assume that $E$ is a topological space, $B$ is closed and $x$ is right-continuous, if any of these assumptions is useful.

Answer 1

Indeed (2) implies (1): if $s \le t$ for some $s \in I$ then $\tau = \inf I \le s \le t$.

The converse is true when $E$ is a topological space, $B$ is closed and $x$ right-continuous (and these assumptions are actually the good assumptions for that, you have simple counter-examples if you remove one assumption).

Indeed, if $\tau \le t$, then for every integer $n \ge 1$, one can choose $t_n \in I$ such that $t \le t_n \le t+1/n$. The right-continuity yields $x(t) = \lim_{n \to +\infty} x(t_n) \in B$, since $B$ is closed and contains every $x(t_n)$.

Answer 2

If $E$ is a topological space, $B$ is closed and $f$ is right-continuous, we can show that either

1. $\tau\in I$; or
2. $I=\emptyset$ and hence $\tau=\infty$.

With this result, it is easy to conclude the desired implication from $(4)$.