# Is the space $B([a,b])$ separable?

Let $a$, $b$ be two real numbers such that $a < b$, and let $B([a,b])$ denote the metric space consisting of all (real or complex-valued) functions $x=x(t)$, $y=y(t)$ that are bounded on the closed interval $[a,b]$ with the metric $d$ defined as follows: $$d(x,y) \colon= \sup_{a\leq t \leq b} \ |x(t) - y(t)|.$$

Then how to determine whether or not this space is separable?

By definition, a metric space $X$ is said to be separable if it has a countable dense subset, that is, if there is a countable subset $M$ of $X$ such that $\bar{M} = X$.

• That would be so kind of you! – Saaqib Mahmood Oct 2 '14 at 4:49
• Saaqib Mahmood, Thank you very much for posting this question. :) – user231343 Sep 16 '18 at 22:39
• @Edi you're welcome. In which context have you found this post to be so useful? Are you a math undergrad student? – Saaqib Mahmood Sep 17 '18 at 12:11
• I've switched from Chemistry to Math and I am learning as much materials I can catch. I didn't expect that B([a,b]) is not separable while C([a,b]) is! – user231343 Sep 17 '18 at 13:54
• @Edi I wish you best of luck in your journey through mathematics and hope you're liking it. – Saaqib Mahmood Sep 19 '18 at 9:35

The space in question is not separable, because you do not have any continuity assumption on the elements. For example, $C([a,b])$ is separable.

To see that your space is not separable, it suffices to construct an uncountable family $(f_i)_i$ in $B([a,b])$ such that $d(f_i, f_j) \geq 1$ for all $i \neq j$ (show this!!, let $(g_n)_n$ be dense in $B([a,b])$, take $\varepsilon = 1/2$ and note that for each $i$ there is some $n_i$ such that $d(f_i, g_{n_i}) < 1/2$. Why does this help you?).

To construct such a family, think for a few minutes or consider the spoiler below.

$$f_{x}\left(y\right)=\delta_{x,y}=\begin{cases}1, & x=y\\0, & x\neq y\end{cases} \text{ for each } x\in[a,b].$$

I leave it to you to check that actually $d(f_x, f_y) = 1$ holds for all $x \neq y$.

• Phoemuex, how to show that the space $C[a,b]$ IS separable? – Saaqib Mahmood Oct 2 '14 at 13:32
• One possibility is to first note that the space of polynomials is dense in $C([a,b])$ (by a theorem of (Stone) Weierstraß). Then observe that each polynomial $p = \sum_{i=0}^n a_i x^i$ can be approximated (uniformly on $[a,b]$) by $\sum_{i=0}^n b_i x^i$ with $b_i \in \Bbb{Q}$. Hence, the countable set of polynomials with rational coefficients is dense in $C([a,b])$. – PhoemueX Oct 2 '14 at 13:37
• Fine. Still my notes ask to prove that space it's separable. – leo Oct 5 '14 at 16:36
• Which one, $C([a,b])$ or $B([a,b])$? In the first case, see the proof sketch in the comment above (or do you have questions?). In the second case, you can not prove it, because it is not separable, as my answer above shows (or do you have questions?) – PhoemueX Oct 5 '14 at 17:13
• We could finalise the proof by observing that for $x\neq y$ we have that$f_x$ and $f_y$ are distance $1$ apart. Then pick for each $x \in [a,b]$ the function $f_x \in B([a,b])$, and use it as the centre of an open ball of radius $1/2$. There are uncountably many non-intersecting balls since the the interval $[a,b]$ is uncountable. Suppose now that the set $D$ is dense in $B([a,b])$. Then since $D$ is dense each open ball must contain an element of $D$, but there are uncountably many balls, hence $D$ must be uncountable. Since any dense subset is uncountable $B([a,b])$ is not separable. – Algebra geek Nov 18 '20 at 22:03