Why is $l^\infty$ not separable?

Solution 1:

For each subset $I$ of the positive integers $\mathbb{N}$, define $e_I\in\ell^\infty$ by $$ (e_I)_i=\begin{cases} 1,&i\in I\,;\\ 0,&i\not\in I\,. \end{cases} $$ Then $d_\infty(e_I,e_J)=1$ whenever $I\neq J$. So $$ \mathcal{B}=\{B(e_I,\frac12):I\subset\mathbb{N}\} $$ is an uncontably infinite collection of disjoint open balls in $\ell^\infty$. Now let $S$ be any dense subset of $\ell^\infty;$ then each ball in the family $\mathcal{B}$ must contain at least one element of $S$ , and these elements must all be distinct, so $S$ must be uncountably infinite. This shows that $\ell^\infty$ is not separable.

Solution 2:

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

Solution 3:

Assume that $A\subset l^\infty$ is countable. We will check that $A$ cannot be dense in $l^\infty$ . Write $A=(a^k)_k$ where $a^k \in l^\infty$ so that $a^k=(a^k_1, a^k_2,\dots)$. For each integer $k$ we define $b_k=a_k^k+1$ if $|a_k^k|\leq 1$ and $b_k=0$ if $|a^k_k|>1$. Note that $b=(b_k) \in l^\infty $ and $|b_k-a^k_k|\geq 1$ for every $k$. Therefore, $\|b-a^k\|_\infty \geq |b_k-a^k_k|\geq 1 $ for every $k$ so $b$ is not in $\overline{ A}$.