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](https://math.stackexchange.com/questions/4704982/is-t-totally-bounded-when-c-ut-is-separable) five days ago, but have not yet received any responses.
Thank you for any suggestions or insights you can provide.



Is $T$ totally bounded when $C_u(T)$ is separable?
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](https://math.stackexchange.com/questions/4704982/is-t-totally-bounded-when-c-ut-is-separable).


Thank you all so much.


