I'm trying to prove the next proposition:
Let $(X_t)_{t\geq0}$ be an adapted process with values in a metric space $(E,d).$ We assume that the sample paths of $X$ are right-continuous, and let $O$ be an open subset of $E.$ Then $T_{O}=\inf\{t\geq 0:X_{t}\in O\}$ is a stopping time of the filtration $(\mathcal{F}_{t+}).$
Because of definition of infimum and inferior bound we have for all $t\geq 0:$ $$\{T_{O}<t\}=\bigcup_{s\in[0,t)}\{X_{s}\in O\},$$ but to ensure measurability we must have a countable measurable union of sets. So my feeling is to prove that $$\{T_{O}<t\}=\bigcup_{r\in[0,t)\cap\mathbb{Q}^{+}}\{X_{r}\in O\},$$ but I'm stuck in this; the last equality have sense to me because of the right continuity, but I'm not sure how to apply it with the open set $O.$
Proving the above finishes the proof.
Any kind of help is thanked in advanced
Let $T_0 <t$ (at a particular point $\omega$). There exists $s<t$ such that $X_s \in O$. Let $\{s_n\}$ be a sequence of rational numbers decreasing to $s$. Since $X_{{s_n}} \to X_s \in O$ it follows that $X_{{s_n}} \in O$ for $n$ sufficiently large. Note that $s_n$ is rational and $s_n <t$. I hope this clears your doubt.