Notions of continuity for stochastic processes I would like to receive some clarification regarding the difference between continuous in probability and continuous almost surely. Using the definition of the wikipedia page (that match the one I have seen in other references), we have
Continuous in probability
for all $\varepsilon>0$
\begin{equation}
\lim_{s\rightarrow t} \mathbb{P}\left(\left\{\omega \in \Omega : |X_s(\omega) - X_t(\omega) | \geq \varepsilon\right\} \right)= 0
\end{equation}
and
Continuous with probability 1 (almost surely)
\begin{equation}
\mathbb{P}\left(\left\{\omega \in \Omega : \lim_{s\rightarrow t} |X_s(\omega) - X_t(\omega) | = 0 \right\} \right)= 1
\end{equation}
Now, I can see that the second condition is stronger than the first one, but to me they seem analogous, since the first should be valid for arbitrarily close $\varepsilon$. Obviously, my intuition is wrong, but I cannot understand how the difference can be significant. Could you please explain how the two definitions give raise to different processes? Thank you!

1. We first examine a much easier variant, i.e., discrete-time process. Let $Y = (Y_n)_{n\in\mathbb{N}_1}$ be a sequence of independent random variables such that
$$ \mathbb{P}(Y_n = 1) = \frac{1}{n} \qquad\text{and} \qquad \mathbb{P}(Y_n = 0) = 1 - \frac{1}{n}. $$
Then it is clear that, for each $\epsilon > 0$,
$$ \lim_{n\to\infty} \mathbb{P}( \left| Y_n - 0 \right| > \epsilon)
\leq \lim_{n\to\infty} \mathbb{P}( Y_n \neq 0)
= \lim_{n\to\infty} \frac{1}{n}
= 0. $$
So $Y_n \to 0$ in probability. However, $\sum_{n=1}^{\infty} \mathbb{P}(Y_n = 1) = \infty$, and so,  by the second Borel-Cantelli's lemma,
$$ \mathbb{P}(Y_n = 1 \text{ infinitely often}) = 1. $$
So it follows that
$$ \mathbb{P}\Bigl(\limsup_{n\to\infty} Y_n = 1\Bigr) = 1
\qquad\text{and}\qquad
\mathbb{P}\Bigl(\liminf_{n\to\infty} Y_n = 0\Bigr) = 1. $$
So $Y_n$ does not converge as $n\to\infty$ with probability one.
Here, the difference between two notions is that the probability $\mathbb{P}(\left|Y_n - 0\right| > \epsilon)$ describes what happens to a 'typical observer $\omega$', and telling that the exceptional event occurs only to a vanishing fraction of observers for each large $n$. However, for a 'fixed observer $\omega$', this exceptional event can happen infinitely often.
2. Now coming back to continuous-time case, we can borrow the above example as follows:
Let $Y=(Y_n)_{n\in\mathbb{N}_1}$ be as above, and define $X = (X_t)_{t\geq 0}$ by
$$ X_t = \begin{cases}
Y_{\lceil 1/t \rceil}, & \text{if } t > 0, \\
0, & \text{if } t = 0.
\end{cases} $$
Then $X_t \to 0$ as $t \to 0$ in probability, whereas $X_t$ does not converge as $t \to 0$ almost surely.
