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The vector space $C_b(\mathbb R)$ of bounded continuous functions on $\mathbb R$ is non-separable: it is possible to produce a direct proof of this fact, mimicking the standard proof for the non-separability of $L^\infty(\mathbb R)$ (reductio ad absurdum: given a sequence $(\phi_n)_{n\in \mathbb N}$ in $L^\infty(\mathbb R)$, construct a function in $L^\infty(\mathbb R)$ at distance 1 of that sequence).

Is it possible to avoid that type of proof and get the non-separability of $C_b(\mathbb R)$ by a more abstract (structural) fact?

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    $\begingroup$ For instance the discrete uncountable metric space $P(\mathbb{N})$ endowed with the Hausdorff distance, embeds isometrically into $C_b(\mathbb{R})$ via $S\mapsto v_S$ s.t. $v_S(x):=\mathrm{dist}_S(x)$. $\endgroup$ – Pietro Majer Feb 10 '15 at 12:33
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    $\begingroup$ What would be an example of a non-direct proof? $\endgroup$ – Ian Morris Feb 10 '15 at 15:06
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    $\begingroup$ This reminds me a little of this question. One sophisticated way to prove this might be that $C_b(\mathbb{R})$ is isometrically isomorphic to $C(\beta \mathbb{R})$ where $\beta \mathbb{R}$ is the Stone-Cech compactification. If $C(\beta \mathbb{R})$ were separable then $\beta \mathbb{R}$ would be second countable (and in fact metrizable), but that isn't true. $\endgroup$ – Nate Eldredge Feb 10 '15 at 19:25
  • $\begingroup$ For what it's worth (if anything), I wouldn't prove $L^\infty(\mathbb{R})$ to be non-separable in the way you suggest: I'd instead do it by observing that the subset $\{\chi_{(-\infty,t)}\colon t \in \mathbb{R}\}$ is isometric to $\mathbb{R}$ with the discrete metric. $\endgroup$ – Ian Morris Feb 11 '15 at 9:40
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Let $\varphi$ be a continuous function supported on $[0,1]$. Then continuum many combinations $\sum c_k \varphi(x+k)$, $c_k\in \{0,1\}$ are separated in our space.

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  • $\begingroup$ Very nice indeed! $\endgroup$ – Transcendental Feb 13 '15 at 2:45
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Consider the functions $f_c(x):=sin(c \pi x)$ for $c\in\mathbb{R}$. If $\frac{c_1}{c_2}\notin\mathbb{Q}$, then $f_{c_1}$ and $f_{c_2}$ have distance 2. This is basically just a restatement of the fact $\{nc + \mathbb{Z} | n\in\mathbb{Z}\}$ is dense in $\mathbb{R}/\mathbb{Z}$ if $c$ is irrational. Since we can find a continuum of real numbers that are pairwise $\mathbb{Q}$-linear independent (even without the full force of the AC), this proves the existence of a continuum-sized discrete subspace of $C_b(\mathbb{R})$.

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Not sure if this is what you are looking for but it is an abstract argument which shows why many of the natural function spaces of bounded functions (continuous, holomorphic, measurable) or operators cannot be separable under the supremum norm. The reason is that each of these spaces has a weaker, natural topology (often known under the name of the strict topology). The latter is the natural one for many applications, in particular, for duality theory. Despite the fact that these do not belong to the standard classes of "nice" locally convex spaces, they do have the property that they satisfy a closed graph theorem for operators with values in a SEPARABLE Banach space (see the first chapter of the monograph "Saks spaces and Applications to Functional Analysis", Prop. I 4.24). Hence if these spaces were separable as Banach spaces, the two structures would coincide, something which only happens in trivial situations. This argument works in the specific situation that you mention but also on a plethora of related examples.

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There is a natural continuous linear surjection $C_b(\mathbb{R}) \to \ell_\infty(\mathbb{Z})$ defined by restricting the function $f \in C_b(\mathbb{R})$ to the subdomain $\mathbb{Z}\subseteq \mathbb{R}$. The space $\ell_\infty(\mathbb{Z})$ is well-known to be non-separable, and a non-separable topological space cannot be the continuous image of a separable one.

To look at the same argument in a different way, define $X$ to be the closed linear subspace of $C_b(\mathbb{R})$ consisting of functions which are affine on each of the intervals $[n,n+1]$. Then $X$ is isometrically isomorphic to $\ell_\infty(\mathbb{Z})$ via the restriction $f \mapsto f|_{\mathbb{Z}}$.

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