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.

I've been grappling with the proof of the "only if" portion for the past three weeks. After numerous unsuccessful attempts, I've even tried to construct counterexamples, suspecting that there is a possible mistake in the statement. Consequently, I am 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," then we can prove the version for $C_u(T,d)$. However, showing $C_b(\overline{T})$ is separable when $C_u(T)$ is separable 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.

I submitted the same query on Mathematics Stack Exchange five days ago, but have not yet received any responses. Thank you for any suggestions or insights you can provide.

Answer 1

What I realized from Mr. Ozawa's comment is as follows:

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.

If there are no serious mistakes, I would like to post the same answer on the original query on Mathematics Stach Exchange.

Thank you all so much.