Prove that two definitions of Separable Processes are equivalent In a number of sources eg (Doob 1950 Stochastic Processes), the definition of a separable stochastic process is as follows:
A random process $ X(t), t \in \mathbb{R} $ is separable if there exists a countable set $ D \subset \mathbb{R} $ and a fixed event $ N $ for which $ \mathbb{P}(N) = 0 $, such that for any closed interval $ B \subset \mathbb{R}$ and open interval $ I \subset \mathbb{R} $ the two sets
    \begin{equation*}
 \left\lbrace \omega: X(t,w) \in B, \text{ for all } t \in I \right\rbrace\text{ and }\left\lbrace \omega: X(t,w) \in B, \text{ for all } t \in I \cap D \right\rbrace
 \end{equation*}
    differ by a subset of $ N. $
Billingsley uses the follow definition:
A random process $ X(t), t \in \mathbb{R} $ is separable if there exists a countable set $ D \subset \mathbb{R} $ such that (with probability 1) for each $ t \in \mathbb{R}$ there exists a sequence $ t_1, t_2, \dots $ such that 
\begin{equation*}
 t_n \in D,\quad  t_n \longrightarrow t, \quad x(t_n) \longrightarrow x(t).
\end{equation*}
The second definition seems much nicer, more intuitive and easier to use. But I'm struggling to show that these definitions are equivalent. 
I know that the second definition is equivalent to supposing that for every open interval $I$ containing $t$, $X(t)$ lies in the closure of $[X(s):s \in I\cap D]$ which looks promising but I'm not sure how to use it.
Any ideas?
 A: The two definitions are not equivalent. Consider the deterministic process:
$X(t)=\left\{\begin{array}{ll}
0 & \text{if }t\notin\mathbb{Q}\\
(-1)^{d(t)}d(t) & \text{if }t\in\mathbb{Q}\text{, where }d(t)\text{ is the denominator of }t\text{ in simplest form.}
\end{array}
\right.$
Let $D=\mathbb{Q}$. Every open interval of $\mathbb{R}$ contains rationals with both odd and even denominators of unbounded size, so for any open interval $I$ the only closed interval $B$  such that $X(t)\in B$ for all $t\in D\cap I$ is $\mathbb{R}$, and so the process is separable on the first definition. On the second definition this process still is separable but not using $D=\mathbb{Q}$ we have to add a countable dense subset of the irrationals. However, the argument that $X$ was separable on the first definition didn't appeal to its behaviour on the irrationals at all so we can have it behave however we like on them and it will still be separable on the first definition. Thus consider the following process where $\omega$ ranges over the probability space $\langle [0,1],\mathscr{Leb},\mathbf{P}_{\mathscr{Leb}}\rangle$ (that is essentially $\omega$ is a uniformly distributed random variable on $[0,1]$): 
$Y(t,\omega)=\left\{\begin{array}{ll}
\frac{\mathbf{1}_{\{t\}}(\omega)}{2} & \text{if }t\notin\mathbb{Q}\\
(-1)^{d(t)}d(t) & \text{if }t\in\mathbb{Q}\text{, where }d(t)\text{ is the denominator of }t\text{ in simplest form.}
\end{array}
\right.$
Here the argument still runs to show that $Y$ is separable by the first definition. However it is not separable by the second definition, since for any countable $D$ with probability 1, $\omega\notin D\cup\mathbb{Q}$, and if so $Y(\omega,\omega)=\frac{1}{2}$ which is a value no sequence in $D$ can converge to, since all the other values of $Y$ are integers.
This suggests that the second definition is stronger than the first, and that is indeed the case. I'll show that for a given $D$ and $\omega^*$ failure of the first condition implies failure of the second: Suppose that for a given $\omega^*$ the condition of the first definition is not satisfied. Then there exists a closed interval $B^*$ and open interval $I^*$ such that $X(t,\omega^*)\in B^*$ for all $t\in I^*\cap D$, but there exists some $t^*\in I\cap D^{*c}$, such that $X(t^*,\omega^*)\notin B^*$. Now consider the second condition applied to $t^*$ we must find sequence $t_n$ in $D$ such that a) $t_n\rightarrow t^*$ and b) $X(t_n,\omega^*)\rightarrow X(t^*,\omega^*)$. But a) implies $t_n$ is in $I^*$ eventually and for all $t\in I\cap D$, $X(t,\omega^*)\in B^*$, so since $B^*$ is closed $\lim_{n\to\infty}X(t_n,\omega^*)\in B^*$ if it exists and so cannot be equal to $X(t^*,\omega^*)$.
Finally if you replace closed interval in the first definition with closed set then the two definitions become equivalent. The same argument works to show that the failure of the first implies failure of the second. Failure of the second for a given $\omega^*$ and $D$ implies failure of the first: Suppose for some $t^*$ no sequence $(t_n)$ in $D$  converging to $t^*$ does $X(t_n,\omega^*)$ converge to $X(t^*,\omega^*)$. Then there exists an open interval $I^*$ containing $t^*$ and an open set $U$ containing $X(t^*,\omega^*)$ such that $t\in I^*\cap D$ implies $X(t,\omega^*)\in U^c$. Therefore $B^*=U^c$ is a closed set such that for all $t\in I^*\cap D$ $X(t,\omega^*)\in B^*$ but it is not the case that for all $t\in I^*$,  $X(t,\omega^*)\in B^*$ (since $t^*\in I^*$).
