# A "too good to be true" claim about separable processes

I am reading the paper . 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 ).

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?

 Asadi et al., Chaining Mutual Information and Tightening Generalization Bounds, 2018. https://arxiv.org/abs/1806.03803
 Van Handel, Probability in high dimensions. https://web.math.princeton.edu/~rvan/APC550.pdf

The inequality $$\lim_{n\to\infty}\mathbb E[\sup_{t\in T}(X_t-X_{\pi_n(t)})]=0 \quad (*)$$ certainly does not follow from separability alone: The indicator of $${\mathbb Q} \cap [0,1]$$ is a deterministic separable process with $$T_0=\Bigl({\mathbb Q} \cup ({\mathbb Q}+\sqrt{2})\Bigr) \cap [0,1]$$. Even seperability and the subgaussian property do not suffice if the metric space is too large. (E.g. consider a Gaussian process indexed by the unit ball $$B$$ of an infinite dimensional Hilbert space, with $$E[(X_t-X_s)^2]=\|t-s\|^2$$ for all $$t,s \in B$$. This process is unbounded in any ball $$B(z,r) \subset B$$.)
However, if you also assume summability of the series (107) in , then $$(*)$$ does follow, and the proof is analogous to the proof of Theorem 5.24 in , using the series (107) instead of the Dudley series, and invoking Theorem 7 of  instead of Lemma 5.1 of .
• Thank you @Yuval for your answer. However I am still confused and cannot see how to easily pass from the proof in  to the conclusion. Could you please detail a bit more? From my understanding, in  they control the reminder $E[\sup_{t\in T}(X_t-X_{\pi_n(t)})]$ by using that it is $0$ if $T$ is finite and $n$ large enough. But to use this argument we would need something like $\lim_{n\to\infty}\sup_k E[\sup_{t\in T^{(k)}}(X_t-X_{\pi_n(t)})]=\sup_k\lim_{n\to\infty} E[\sup_{t\in T^{(k)}}(X_t-X_{\pi_n(t)})]=0$, where $T^{(k)}$ are finite subset whose union is $T_0$. But how to show it?
• Moreover I don't see how the series (107) helps in solving the question. The random variable $W$ does not appear in (*), so we might choose any random variable like that which is independent of the process, and all the mutual information would always be $0$, making the series (107) summable.
• The series (107) helps, with this $W$ in view of Theorem 7 of . Take $T^{(k)}$ as the first $k$ elements of $T_0$ in some enumeration. Jan 26 at 17:47