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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?

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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
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@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
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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 $$f_{A}\left(x\right):=\arctan\left(dist\left(\rho\left(x\right),A\right)\right)\cdot e$$ $\left(x\in O\right).$

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@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
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