What is the completion of $C_b([0,1] \cap \mathbb{Q})$?

Start with the space $Z = [0,1] \cap \mathbb{Q}$ equipped with the standard (metric) topology. Consider the space $C_b(Z)$ of all bounded continuous complex-valued functions on $Z$. It seems to me that the ordinary sup norm is well-defined this space, but $C_b(Z)$ with the sup norm is not complete. What can we say about its completion?

• $C_b(Z)$ has a lot of functions with no continuous extension to $[0,1]$, so its completion is going to be much bigger than $C([0,1])$, I think.
• The completion ought to be a C*-algebra, right? If so, what is its Gelfand dual? It would have to be bigger than $\beta Z$, I'd think?

More generally, starting with a dense subset $Z$ of a compact Hausdorff $X$, what is the relationship between the completion of $C_b(Z)$ and $X$? Can we view it as a space of functions on some sort of space of measures on $X$, perhaps?

Solution 1:

Let $X$ and $Y$ be metric spaces. The completeness of $C_b(X;Y)$ is related only to the completeness of $Y$. In particular, if $Y$ is complete, then $C_b(X,Y)$ is complete with $d_\infty(f,g):=\sup_{x\in X} d_Y(f(x),g(x))$. See e.g. here when $Y=\mathbb R$. The case $Y=\mathbb C$ is identical, and the proof can easily be adapted for general complete $Y$. This is because the uniform distance $d_\infty$ (or the uniform norm if $Y=\mathbb R$ or $\mathbb C$) only depends on the distance on $Y$.

What you might have in mind in the rest of the question is that $C_b(Z)$ is actually isomorphic to $C(\beta Z)$. This is true by the universal property of the Stone–Čech compactification since $Z$ is completely regular Hausdorff. (See e.g. [A] Cor. 4.2.5). It holds in general for dense subsets of compact Hausdorff spaces as well, since they are in particular completely regular Hausdorff (see here).

Note: this property of $\beta Z$ has nothing to do with the completeness of $C_b$-spaces.

[A] V. Runde, A taste of topology, Springer 2005