If $u$ is rational, can we show $\inf\{t:x(t)\in B\}\le u$ iff $x(s)\in B$ for some rational $s\le u$?

Question

Let $(E,d)$ be a metric space, $B\subseteq E$ be closed and nomepty, $x:[0,\infty)\to E$, $I:=\{t\ge0:x(t)\in B\}$ and $\tau:=\inf I$. If $I$ is nonempty and $\tau\in I$, then we easily see$^1$ that $$\tau\le t\Leftrightarrow\exists s\in I:s\le t\tag1$$ for all $t\ge0$. Moreover, if $I$ is nonempty and $x$ is right-continuous, then $\tau\in I$.

Are we able to show that if $t\in[0,\infty)\cap\mathbb Q$, then $$\tau\le t\Leftrightarrow\exists s\in I\cap\mathbb Q:s\le t?\tag2$$

We may clearly use that $\mathbb Q$ is dense in $\mathbb R$, but I'm still not sure how to argue. If necessary, feel free to assume that $x$ is right-continuous.

---

$^1$ "$\Rightarrow$": Assume the contrary. Then $$\forall s\in I:s>t\tag3$$ and hence $t$ is a lower bound for $I$. Thus, $$t\le\tau$$ by definition of the infimum and hence $$t=\tau\in I;$$ in contradiction to $(3)$.

"$\Leftarrow$": Let $s\in I$ with $s\le t$. Since $\tau$ is a lower bound for $I$, $$\tau\le s\le t.$$

Answer 1

It occurs to me to do this:

Let $I:=\{t\ge0:x(t)\in B\}$ such that $I\cap \mathbb Q\neq \emptyset$ and $\tau:=\inf I$, I saw that you want to use the continuity of x so you can use it to see: $$\tau:=\inf I\Rightarrow \exists (t_n)\in [0,\infty)\text{ such that } t_n\to\tau \Rightarrow x(t_n)\to x(\tau)\in B,\text{ since } B \text{ is closed}.$$ In other words $\tau=\min I$. Then we want to demonstrate $$\tau\leq u, (\text{with }u\in\mathbb Q)\;\text{ iff }\;x(s)\in B\text{ for some rational }s\leq u$$ or equivalently $$\tau> u, (\text{with }u\in\mathbb Q)\;\text{ iff }\;x(s)\not\in B\text{ for all rational }s\leq u$$ $(\Rightarrow)$ Suppose as a hypothesis that $\tau>u$. Suppose as an auxiliary hypothesis $x(s)\in B$ for some rational $s\leq u(<\tau)$. Which is a contradiction to the infimo.

$(\Leftarrow)$ Since $$x(s)\not\in B\text{ for all rational }s\leq u\Rightarrow x(s)\not\in B\text{ for all real }s\leq u$$ This is for the continuity of $x(t)$. Then $x>u$, therefore $\tau\geq u$. But since $u\leq u$ is rational we have that $x(u)\not\in B$. Since $\tau$ is the minimum then $\tau>u$.

Answer 2

Let $(E,d)$ be $[0,\infty)$ with the Euclidian metric. Let $B=\{0\}$. Let us consider any fixed irrational number in $[0,\infty)$ say $\sqrt{2}$. Define the function $x:[0,\infty) \to [0,\infty)$ as follows,

$x(r)=\begin{cases}r-\sqrt{2}, \text{ if } r \geq \sqrt{2} \\ 1, \text{ if } r < \sqrt{2} \end{cases}$

It can be easily verified that $x$ is right continuous.

As per your definition, $\tau=\inf I=\inf\{t \geq 0: x(t) \in B\}=\inf\{t \geq 0: x(t)=0\}=\inf\{\sqrt{2}\}=\sqrt{2}$. Now, consider any rational number $t \in [0,\infty)$ which is greater than $\sqrt{2}$. We have $\tau < t$. But since $I$ is a singleton set containing an irrational number, the conclusion $\exists s\in I \cap \mathbb{Q} : s \leq t$ is false.

The issue with the argument you gave for the forward direction is that, $\forall s \in I, s>t$ need not be true. This happens for example when $I$ does not contain any rational number, as in the counterexample given above.