0
$\begingroup$

Consider the metric space $\mathcal{C}(\mathbb{R})$, the set of all compactly supported continuous functions defined from real line to itself and the distance function induced by the supremum norm. We know that this metric space is separable because the polynomials with rational coefficients will form a countable dense subset of this metric space. Now, instead of compactly supported continuous functions, let us consider the space of all continuous and bounded functions defined from real line to itself and the norm is the usual supremum norm. Let denote this metric space by $(\mathcal{C}_b(\mathbb{R}),||.||_{\infty})$. Will this metric space be separable?

$\endgroup$
2
$\begingroup$

No.

Let $\lbrace f_n \,|\, n \in \mathbb{N}\rbrace$ be any countable set of bounded continuous functions on $\mathbb{R}$. For every positive integer $n$, pick a real number $y_n \in [-1,1]$, such that $|f_n(n) - y_n| \geq 1$. (For instance $y_n:=-1$ if $f_n(n) \geq 0$ and $y_n:=1$ if $f_n(n) \leq 0$.) Define a bounded, piecewise linear function $g : \mathbb{R} \rightarrow \mathbb{R}$ by imposing $g(n):= y_n$ for all $n$. Then $||g-f_n||_\infty \geq 1$ for all $n \in \mathbb{N}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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