Why is $l^\infty$ not separable?

Question

My functional analysis textbook says

"The metric space $l^\infty$ is not separable."

The metric defined between two sequences $\{a_1,a_2,a_3\dots\}$ and $\{b_1,b_2,b_3,\dots\}$ is $\sup\limits_{i\in\Bbb{N}}|{a_i-b_i}|$.

How can this be? Isn't the set of sequences containing complex numbers with rational coefficients the required countable dense subset of $l^\infty$?

Thanks in advance!

Answer 1

Check that the sequences with only $0's$ and $1's$ have dist $1$, and they are uncountable. Can you now finish the problem?

Answer 2

A rephrasing of the solutions offered:

The idea is that the unit ball in $l ^{\infty}$ , i.e., the set of points in $l^{\infty}$ of norm $1$ under the $Sup A_n$ norm, does not have a countable dense subset. Consider any two such points $a_i, a_j ; i\neq j, d(a_i, a_j):=Sup|a_i -a _j| \geq 1 $ . Therefore a ball $B(a_i, 1/2)$ will not intersect any $a_j$. There are uncountably-many such points and they do not intersect each other. If there was a dense countable subset $D$ , it would have to contain one element each , so $D$ could not be countable.