I am reading the paper [1]. At page 18, eq 115, it is claimed the following:

Given a separable process $(X_t)_{t\in T}$, we have $\lim_{n\to\infty}\mathbb E[\sup_{t\in T}(X_t-X_{\pi_n(t)})]=0$.

Here the only hypothesis on the labelling space $T$ is that it is a bounded metric space. $(X_t)$ is actually assumed to be a subgaussian process, but this fact should not matter for the property stated, or at least the inequality is justified only using the separability in the paper. $\pi_n$ defines a sequence of refining nets. This means that $d(\pi_n(t), t)\leq \varepsilon_n$ for all $t$ and all $n$ (where $d$ is the distance defined on $T$ and $\varepsilon_n$ is a vanishing decreasing positive sequence), and $\pi_n(T)$ is a finite set.

The definition used for separability (Definition 5 in the paper) is the following:

A random process $(X_t)_{t\in T}$ is said separable if there is a countable set $T_0\subseteq T$, dense in $T$, such that $$\mathbb P\left(X_t\in\lim_{s\to t, s\in T_0}X_s, \quad\forall t\in T\right) = 1\,.$$

In the definition above $x\in \lim_{s\to t, s\in T_0}x_s$ means that there is a sequence $x_s$ in $T_0$ which tends to $x$.

Does the separability of a process actually guarantee such a strong convergence of $X_{\pi_n(t)}$ to $X_t$? What strikes me the most is that the separability involves only the existence of a dense subset, which might have nothing to do with the support of the nets $\pi_n(T)$.

The justification of the property given in the paper is

(see proof of Theorem 5.24 in [2]).

However, looking at the referenced proof I could not find anything suggesting that $\lim_{n\to\infty}\mathbb E[\sup_{t\in T}(X_t-X_{\pi_n(t)})]=0$.

Am I actually missing something obvious?

[1] Asadi et al., Chaining Mutual Information and Tightening Generalization Bounds, 2018. https://arxiv.org/abs/1806.03803

[2] Van Handel, Probability in high dimensions. https://web.math.princeton.edu/~rvan/APC550.pdf