The space of bounded continuous functions are not separable

Let $C_b(\mathbb{R})$ be the space of all bounded continuous functions on $\mathbb{R}$, normed with $$\|f\| = \sup_{x\in \mathbb{R}} \|f(x)\|$$ 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?

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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 9 '12 at 16:06
Can you expand? how will the tent-function work? – Johan Dec 9 '12 at 16:32
Do you know that $\ell^{\infty}$ is not separable? It could be helpful when proving this claim. – T. Eskin Dec 9 '12 at 16:33

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 $C_b(\Bbb R)$.

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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 9 '12 at 16:29
@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 9 '12 at 16:33
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 9 '12 at 16:35
@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 9 '12 at 16:38
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 9 '12 at 20:21

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 $t_1\neq t_2$, $$\lVert f_{t_1}-f_{t_2}\rVert_\infty=1.$$

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Who should I accept when I get 2 good answere and the difference is 2 seconds? – Johan Dec 9 '12 at 17:30
Perhaps I'm missing something, but shouldn't one consider real-valued functions here? – David Mitra Dec 9 '12 at 17:35
@DavidMitra Yes, but I think it makes the computations a little more easier. – Davide Giraudo Dec 9 '12 at 17:40

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.

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