2

It's well-known that

  • Theorem 1 A metric space is compact IFF its space of bounded, continuous, real-valued functions is separable in the uniform topology.

I would like to prove its complementary result, i.e.,

Theorem 2 Let $(X, d)$ be a separable metric space. Then the space of all real-valued bounded uniformly continuous functions on $X$ is separable.

Could you have a check on below proof?

My attempt We need the following lemmas, i.e.,

  • Lemma 1: If $X$ is separable, then there exists a compact metric space $Y$ and a map $T: X \to Y$ such that $T$ is a homeomorphism from $X$ onto $T(X)$.
  • Lemma 2 Suppose $Y$ and $Z$ are metric spaces, and $Z$ is complete. Also suppose $X$ is a dense subset of $Y$, and $f: X \rightarrow Z$ is uniformly continuous. Then $f$ has a uniquely determined extension $\bar{f}: Y \rightarrow Z$ given by $$ \bar{f}(y)=\lim _{\substack{x \rightarrow y \\ x \in X}} f(x) \quad \text { for } y \in Y, $$ and $\bar{f}$ is also uniformly continuous. [Theorem 2.1 at page 10 of Amann/Escher's Analysis III]
  • Lemma 3 Let $f:X \to Y$ be continuous and $X,Y$ be metric spaces where $X$ is compact. Then $f$ is uniformly continuous.

By Lemma 1, there is a metric $d'$ on $X$ such that $d,d'$ induce the same topology and $d'$ is totally bounded. Let $X_1$ be the completion of $X$ w.r.t. $d'$. Then $(X_1, d')$ is compact.

By Theorem 1, the space $\mathcal C_b(X_1)$ of all real-valued bounded continuous functions on $X_1$ is separable. Let $\mathcal{UC}_b (X)$ be the space of all real-valued bounded uniformly continuous functions on $X$.

By Lemma 3, $\mathcal{UC}_b (X_1) = \mathcal{C}_b (X_1)$. By Lemma 2, $\mathcal{UC}_b (X)$ can be embedded isometrically into $\mathcal{UC}_b (X_1)$. This completes the proof.

Update: It seems I made a fatal mistake. It should be

"$\mathcal{UC}_b (X, d')$ can be embedded isometrically into $\mathcal{UC}_b (X_1)$"

instead of

"$\mathcal{UC}_b (X)$ can be embedded isometrically into $\mathcal{UC}_b (X_1)$".

Of course, $\mathcal{UC}_b (X, d') \neq \mathcal{UC}_b (X, d)$.

Analyst
  • 6,351
  • I have not read your proof. But "Theorem 2" is false. Suppose $X$ is the closed unit ball in the real Banach space $\ell_1. $ Then the space of real, bounded, uniformly continuous functions on $X$ has a non-separable subspace consisting of the closed unit ball of $\ell_{\infty}$. – DanielWainfleet Nov 29 '22 at 19:06
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    @DanielWainfleet Thank you so much for your counter-example. Could you post it as an answer? – Analyst Nov 29 '22 at 19:13
  • @DanielWainfleet Is ${f_{\restriction X} : f \in B}$ with $B$ the closed unit ball of $\ell_{\infty}$ the non-separable subspace you mentioned? – Analyst Nov 29 '22 at 19:30
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    Yes, to be precise about it, the members of the subspace are restrictions to the domain $X$ of the members of the closed unit ball of $\ell_{\infty}.$ – DanielWainfleet Dec 05 '22 at 08:45
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    BTW if Theorem 2 were true then by Theorem 1 every separable metric space would be compact, which is absurd. A more elementary counter-example: $X=\Bbb R$ and $d(x,y)=|x-y|.$ For $n\in \Bbb Z$ let $f_n(x)=(x-2n-1)^2(x-2n+1)^2$ if $x\in [2n-1,2n+1]$ and let $f_n(x)=0$ if $x\in\Bbb R\setminus [2n-1,2n+1].$ Now for any $S\subseteq \Bbb Z$ let $f^S(x)=f_n(x)$ if $n\in S$ and $x\in [2n-1,2n+1],$ otherwise $f^S(x)=0.$ If $S,T$ are unequal subsets of $\Bbb Z$ and $n\in S\Delta T$ then ${f^S(2n),f^T(2n)}}=$ ${0,1}$. – DanielWainfleet Dec 05 '22 at 09:08

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