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Let $C_b(\mathbb{R})$ be the space of all bounded continuous functions on $\mathbb{R}$, normed with $$\|f\| = \sup_{x\in \mathbb{R}}{\lvert f(x)\rvert}$$ Show that the space $C_b(\mathbb{R})$ is not separable.

A space is separable if there is a dense countable subset. How do I prove that something is not separable? would it matter if we looked at $C_b([0,1])$ instead?

davyjones
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Johan
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    Maybe you can try the tent-functions with different centers. Note that $C_b([0,1])=C([0,1])$, which is separable by Stone-Weierstrass. – Hui Yu Dec 09 '12 at 16:06
  • Can you expand? how will the tent-function work? – Johan Dec 09 '12 at 16:32
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    Do you know that $\ell^{\infty}$ is not separable? It could be helpful when proving this claim. – T. Eskin Dec 09 '12 at 16:33

3 Answers3

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Consider the subset $K$ of $C_b(\Bbb R)$ consisting of functions that are either $0$ or $1$ at the integers. There is an uncountable subset $S$ of $K$ such that: $$\tag{1}\Vert x-y\Vert\ge 1,\ \ \text{whenever}\ \ x,y\in S\ \text{with}\ x\ne y.$$

Now, given a countable subset $B$ of $C_b(\Bbb R)$ it follows from $(1)$ that there is an $s\in S$ that is distance at least $1/2$ from every element of $B$. Then $B$ is not dense in $C_b(\Bbb R)$.

David Mitra
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    Do you mean $f_\alpha(x) = 1$ only at $x = \alpha $ and zero otherwise? Is such functions rally continuous? – Johan Dec 09 '12 at 16:29
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    @Johan. No, I meant simply that $f$ is bounded and continuous and $f(n)$ is either $0$ or $1$ for every integer $n$. – David Mitra Dec 09 '12 at 16:33
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    I understand, but how do we know there is an uncountable subset? with that property? There seems to be at least $2^{\mathbb{N}}$ to combine this... is that uncountable? – Johan Dec 09 '12 at 16:35
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    @Johan So, given a sequence $(b_n)$ of zeroes and ones, define $f(n)=b_n$ for each $n$ and define the rest of $f$ by "connecting the dots". You can take $S$ to be the set of these functions. This is uncountable since there are uncountably many infinite binary sequences. Two distinct members $x$ and $y$ of $S$ will have different values at some integer, so then $\Vert x-y\Vert\ge 1$ . – David Mitra Dec 09 '12 at 16:38
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    A general principle that’s in the background here: to show that a space is non-separable, it is enough to define an uncountable family of pairwise disjoint open sets. (In this case, the balls of radius 1/2 around the functions in K.) This is a very useful go-to approach for proving non-separability. – Peter LeFanu Lumsdaine Dec 09 '12 at 20:21
  • This answer almost seems like it's invoking the axiom of choice if we want to argue that the $s$ exists as stated in this solution. If we take the map $\psi : K \to {0,1}^{\mathbb Z}$ given by $\psi(f) = (f(n)){n \in \mathbb Z}$, then $\psi$ is surjective, so we choose $f\omega \in \psi^{-1}({\omega})$ for each $\omega \in {0,1}^{\mathbb Z}$. I don't see any other way to construct this set $S$. Am I right about this?

    (Of course there's nothing wrong with the axiom of choice, but I'm wondering if there's a less "nuclear" option for this problem.)

     – D Ford Jan 26 '21 at 01:43
  • To see why the comment of @PeterLeFanuLumsdaine is valid: https://math.stackexchange.com/questions/2137105/how-to-prove-a-space-is-not-separable#:~:text=Let%20(Ui)i%E2%88%88,ni%E2%88%88Ui. – XXX Apr 26 '21 at 10:10
  • @DFord You don't need to use the axiom of choice. Take a continuous, compactly supported function $\phi$ such that $(\psi(\phi))n=\delta{0,n}$ (i.e., $\phi(0)=1$ and $\phi(n)=0$ for $n\in\mathbb Z\setminus{0}$); you can easily build it explicitly as you wish. Then you can define your $f_\omega$ as $\sum_{n\in\mathbb Z}\phi(x-\omega_n)$. It is not difficult to show that $f_\omega$ is well-defined, continuous and $\psi(f_\omega)=\omega$. – Lorenzo Pompili Aug 05 '22 at 20:17
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Here is also one way, incase you know that $\ell^{\infty}$ is not separable, the set of bounded sequences with sup-norm that is.

Define $\Lambda:C_{b}(\mathbb{R})\to\ell^{\infty}$ by $f\mapsto (f(n))_{n=1}^{\infty}$. By choosing any $(x_{n})_{n=1}^{\infty}\in \ell^{\infty}$ we can define a continuous bounded function $f\in C_{b}(\mathbb{R})$ by setting $f(n)=x_{n}$ for all $n\in\mathbb{N}$ and extending it to $\mathbb{R}$ by connecting the imagepoints of each natural number linearly. We may take $f(x)= x_{1}$ for all $x<1$, for example. Since $f\mapsto (x_{n})_{n=1}^{\infty}$, then $\Lambda$ is a surjection. It is also continuous, because \begin{equation*} \|\Lambda(f)-\Lambda(g)\|_{\infty}=\sup_{n\in\mathbb{N}}|f(n)-g(n)|\leq \sup_{x\in\mathbb{R}}|f(x)-g(x)|=\|f-g\|_{\infty} \end{equation*} for all $f,g\in C_{b}(\mathbb{R})$.

Now if $C_{b}(\mathbb{R})$ was separable, it would have a countable and dense subset $\mathscr{D}$. Since the image of a dense subset under a continuous surjection is dense, we have that $\Lambda(\mathscr{D})$ is a countable, dense subset of $\ell^{\infty}$. This would imply that $\ell^{\infty}$ is separable, which is a contradiction.

T. Eskin
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Consider the maps $f_t(x):=\exp(itx)$. We have, if $t_1\neq t_2$, that \begin{align*} |f_{t_1}(x)-f_{t_2}(x)|&=|\exp(it_1x)-\exp(it_2x)|\\ &=\frac 12\left|\exp\left(i\frac{t_1+t_2}2x\right)\right|\cdot \left|\exp\left(i\frac{t_1-t_2}2x\right)-\exp\left(-i\frac{t_1-t_2}2x\right)\right|\\ &=\sin\left(\frac{t_1-t_2}2x\right), \end{align*} which gives, when $t_1\neq t_2$, $$\lVert f_{t_1}-f_{t_2}\rVert_\infty=1.$$

Davide Giraudo
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