Is $T$ totally bounded when $C_u(T)$ is separable?

Question

I'm seeking help with a question regarding the space of bounded and uniformly continuous functions $C_u(T,d)$, where $(T,d)$ is a metric space. In this context, $C_u(T)$ is a closed subspace of $C_b(T)$, therefore it is a Banach space as well.

In the second edition of Giné and Nickl's Mathematical Foundations of Infinite-Dimensional Statistical Models (2021), page 17 , a statement reads,

The Banach space $C_u(T,d)$ is separable if (and only if) $(T,d)$ is totally bounded.

While I've been struggling with the proof of the "only if" part for the past three weeks, I've also attempted to construct counterexamples. I'm now seeking additional ideas or guidance to approach this problem. Here is what I have tried so far:

1. I proved $C_b(T,d)$ is separable if and only if $(T,d)$ is compact, following Conway's A Course in Functional Analysis (2007) Theorem V.6.6 (p.140). If we can prove "when $C_u(T,d)$ is separable, the completion of $T$, denoted as $\overline{T}$, is compact," we can prove the version for $C_u(T,d)$. However, obtaining the Banach space isomorphism $C_u(T)\simeq_{\mathrm{Ban}}C_b(\overline{T})$ proves challenging, although $C_u(T)\simeq_{\mathrm{Ban}}C_u(\overline{T})$ is a relatively straightforward result.
2. Abandoning the use of the $C_b(T)$ version's result, I examined other paths. If $C_u(T)$ is separable, the closed unit ball of the dual space $B^*\subset C_u(T)^*$ is metrizable. Coupled with the fact $B^*$ is always $w^*$-compact, we find $B^*$ to be $w^*$-sequentially compact. Hoping this would offer a proof to the original problem, I turned to the property of a metric space being totally bounded if and only if every sequence has a Cauchy subsequence. For an arbitrary sequence $\{x_n\}\subset T$, we obtain a $w^*$-convergent subsequence $\{\delta_{x_{n_k}}\}\subset B^*$. However, attempts to prove $\{x_{n_k}\}\subset T$ as a Cauchy sequence by evaluating at some specially constructed $f_1,f_2,\cdots\in C_u(T)$ have been unsuccessful.

Thank you for any suggestions or insights you can provide.

Update: I also posted the same query on mathoverflow

Answer 1

I received an incredibly clear and simple proof idea to the same query I posted on mathoverflow. I'll present it here in my own terms.

Step1. There exists an $\epsilon>0$ and a sequence $\{x_n\}\subset T$ such that the open balls $\{U_\epsilon(x_n)\}_{n\in\mathbb{N}}$ are disjoint.

This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup_{n=0}^NU_\epsilon(x_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup_{n=0}^NU_\epsilon(x_n)$, we have $$d(x,\{x_n\}_{n=0}^N)<2\epsilon.$$ This implies that the family $$\{U_{2\epsilon}(x_n)\}_{n=0}^N$$ covers $T$, contradicting the assumption that $T$ is totally bounded.

Step2. There is a Banach space embedding $l^\infty\hookrightarrow C_u(T)$.

By defining $f_n(x):=\max\left\{1-\frac{d(x,x_n)}{\epsilon},0\right\}$ for all $n\in\mathbb{N}$, there exist functions $\{f_n\}\subset C_u(T;[0,1])$ such that $$f_n(x_n)=1,\quad f_n|_{T\setminus U_\epsilon(x_n)}=0,\quad n\in\mathbb{N}.$$ Using these, we can define the following bounded linear operator: $$(a_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}a_nf_n$$ This operator preserves the norm, so it is also a Banach space embedding.

Step3. $l^\infty$ is not separable.

The set of indicator functions $\{\chi_I\}_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.

Separable spaces do not have a non-separable subspace, so this constitutes a contradiction.