3

Let C$_u$(X) represents the space of uniformly continuous functions from (X,d$_1$) to (R,|·|).

I have met with a conclusion that C$_u$(X) is separable if X is totally bounded

Can anyone prove this theorem?

ze min jiang
  • 313
  • 1
    This is not true: $X=\mathbb{R}$ is not totally bounded, but $C_u(\mathbb{R},\mathbb{Q})$ is separable. – Luiz Cordeiro Sep 19 '19 at 04:45
  • 1
    Uniformly continuous maps from the real line to itself need not be bounded, so the $L^{\infty}$ norm does not even make sense. – Kavi Rama Murthy Sep 19 '19 at 05:45
  • @LuizCordeiro Sorry I'm wrong. I have edited my question. – ze min jiang Sep 19 '19 at 11:16
  • 2
    The completeion $\tilde X$ of $X$ is compact and $C_u(X)$ is isomorphic to $C(\tilde X)$ -- the restriction map $C(\tilde X)\to C_u(X)$, $f\mapsto f|_X$ is an isomorphism. All you need to know is thus the separability of $C(X)$ for compact metric spaces. – Jochen Sep 19 '19 at 11:28
  • @Jochen Thanks! Nice idea. https://math.stackexchange.com/questions/1331321/cx-is-separable-when-x-is-compact This site solves the following problem. – ze min jiang Sep 19 '19 at 11:37
  • @zeminjiang Using the idea above (completing $X$ to a compact), this is a classic application of Stone-Weierstrass: Take a countable dense subset $D$ of $X$ and for every $y\in D$ and every $q>0$ rational, a continuous function $f_{y,q}$ such that $f_{y,q}(x)=1$ if $d(x,y)\leq q/2$ and $f_{y,q}(x)=0$ if $d(x,y)>q$ (e.g. $f_{d,y}(x)=\max(0,\min(1,2-{2}\frac{d(x,y)}{q}))$). The $\mathbb{Q}$-algebra generated by the $f_{d,y}$ is countable and dense by Stone-Weierstrass. – Luiz Cordeiro Sep 19 '19 at 14:43
  • What topology does $C_u(X)$ have? You don't menition it anymore and sup-norm is not available. So compact-open, or pointwise topology? – Henno Brandsma Sep 19 '19 at 21:28
  • @Hennessy It has the topology generated by sup-norm. The total boundedness makes the sup-norm available since it's always bounded. – ze min jiang Sep 20 '19 at 14:43
  • @LuizCordeiro It's true. Thanks for your answer. I used to see a similar construction in the proof of C[0,1]'s separability. But I fail to deal with those overlapping parts in the general cases. This theorem helps me avoid those complex parts. – ze min jiang Sep 20 '19 at 14:48
  • @Jochen you want to take bounded uniformly continuous functions on $X$ for this to be an isomorphism. – Jakobian Dec 13 '22 at 14:47
  • @Jakobian For totally bouded metric spaces $X$, every uniformly continuous real-valued function is bounded. – Jochen Dec 14 '22 at 08:11
  • @Jochen ahh, I see, thanks. – Jakobian Dec 14 '22 at 18:22

0 Answers0