When $X_sWhen I have shown, for $s\le t$ and for two continuous stochastic process an inequality:
$$ X_s \le Y_t$$ P-a.s.
How can I deduce that this P-a.s. simultaneously for all rational $s\le t$ ?
Thank you for your help
EDIT: According to Ilya's answer, I see that we have $$P(X_s\le Y_t\text{ simultaneously for all rationals }s\le t) = 1.$$ 
How could we use continuity of $X,Y$ to deduce $P(X_s\le Y_t,s\le t)=1$. Of course we take sequences of rational, however I mess up the details. So a detailed answer how to do this, would be appreciated.
 A: If I got you correct, you have $P(X_s\leq Y_t) = 1$ for all $s\leq t$. Recall that if $P(A_n) $ for all $n\in \mathbb N$ then also you have $P(\bigcap_n A_n) = 1$ since
$$
P(\bigcap_n A_n) = 1-P(\bigcup_n A^c_n)\geq 1-\sum\limits_n P(A^c_n) =1.
$$
Now take $s_n$ be the $n$-th rational number that is less or equal to $t$ and $A_n=\{X_{s_n}\leq Y_t\}$. You can even use continuity of $X,Y$ to show that $P(X_s\leq Y_t,s\leq t) = 1$.
To show the latter, consider two sets: 
$$
  C = \{\omega\in \Omega: X_t(\omega),Y_t(\omega)\text{ are continuous }\}
$$
$$
  D = \{\omega\in \Omega:X_s(\omega)\leq Y_t(\omega) \text{ simultaneously for all rational }s\leq t\}.
$$
It is given to us that $\mathsf P(C) = 1$ and we have proved above that $\mathsf P(D) = 1$. As a result, 
$$
  \mathsf P(C\cap D) = 1-\mathsf P(C^c\cup D^c)\geq 1-\mathsf P(C^c)-\mathsf P(D^c) = 1.
$$
Consider now any $\omega\in C\cap D$. 


*

*It holds that for this $\omega$: $X_s(\omega)\leq Y_t(\omega)$ for all rational $s\leq t$. Since both $X,Y$ are continuous on $\omega$, it follows that $X_s(\omega)\leq Y_t(\omega)$ for all $s\leq t$. Indeed, if that would not be true, i.e. for some $s'\leq t$ you have $X_{s'}(\omega)>Y_{t}(\omega)$ then $X_s(\omega)>Y_t(\omega)$ in the neighborhood of $s'$ which cannot happen since there will be at least one rational number $s$.

*Thus, for any $\omega\in C\cap D$ the desired relation holds: $C\cap D \subseteq \{X_s\leq Y_t,s\leq t\}$, so that
$$
  1 =\mathsf P(C\cap D) \leq \mathsf P\{X_s\leq Y_t,s\leq t\}.
$$
A: This follows from the fact that the complement of the event $[\forall s\leqslant t,\,X_s\leqslant Y_t]$ is the event
$$
\left[\exists s\leqslant t,\,X_s\gt Y_t\right]\ =\left[\exists n\in\mathbb N,\,\exists s\in\mathbb Q,\,\exists t\in\mathbb Q,\,s\leqslant t,\,X_s\geqslant Y_t+\frac1n\right],
$$
hence
$$
\left[\exists s\leqslant t,\,X_s\gt Y_t\right]\ =\bigcup\limits_{s\leqslant t,\, s\in\mathbb Q,\,t\in \mathbb Q}\ \bigcup_{n\geqslant1}\ \left[X_s\geqslant Y_t+\frac1n\right].
$$
Since $\mathrm P(X_s\leqslant Y_t)=1$ for every $s\leqslant t$, $\mathrm P(X_s\geqslant Y_t+\frac1n)=0$ for every $n\geqslant1$. The union on the RHS of the displayed identity above is countable hence $\mathrm P(\exists s\leqslant t,\,X_s\gt Y_t)=0$. Considering the complement, one gets
$$
\mathrm P(\forall s\leqslant t,\,X_s\leqslant Y_t)=1.
$$
