MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $O$ be an open subset of the separable Hilbert space $H.$ Let $E$ be a separable Banach space. Is it true that $C^0_b(O;E),$ the space of bounded continuous maps $O\rightarrow E$, endowed with the $C^0$-norm, is separable? If YES, where can I find I proof of this fact?

share|cite|improve this question
Is $C^0$ the uniform norm? If so, the answer is clearly no. You can find an uncountable set such that the distance between any two is 1; for each set of integers pick a continuous function which is 1 on those integers and 0 on the rest. – Nate Eldredge Feb 1 '11 at 23:06
Er, here I am thinking of taking $O = H = \mathbb{R}$ and $E = \mathbb{R}$ as well. – Nate Eldredge Feb 1 '11 at 23:13
@Nate: I am confused. The space $C^0_b(\mathbb{R})$ is separable! The proof is simple - it follows from the separability of $C^0([0,1]).$ Fix a $\delta>0$ sufficiently small and cover $\mathbb{R}$ by intervals of the form $[n-\delta,n+1+\delta].$ On each of these there is a countable dense set. Now glue all these together on the overlapping parts to make the resulting function continuous. This is how you arrive at your countable dense set! – Orbicular Feb 2 '11 at 8:30
@Orbicular: With countably many choices on each of countably many intervals, there seem to be continuum many things you can get by gluing. – Andreas Blass Feb 2 '11 at 9:53
@Orbicular: But don't you end up "gluing a countable set to itself countably many times"?? That doesn't give a countable set: think about the cardinality of $\{0,1\}^{\mathbb N}$. – Matthew Daws Feb 2 '11 at 9:54
up vote 2 down vote accepted

The answer is negative. For, pick some non-zero $e$ in $E$, and choose a surjection $\rho\in C\left(O,\mathbb{R}\right)$ (there exists !).

Next, consider the (uncountable, uniformly discrete) family of functions { $f_{A}$; $A\subset\mathbb{Z}$ nonempty } $\subset C_{b}^{0}\left(O,E\right)$, expressed by $$f_{A}\left(x\right):=\arctan\left(dist\left(\rho\left(x\right),A\right)\right)\cdot e$$ $\left(x\in O\right).$

share|cite|improve this answer
@all the guys: You are absolutely right, sorry! I was just not seeing things clearly... Thanks! Now at least I understand those things... – Orbicular Feb 2 '11 at 14:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.