Why is $l^\infty$ not separable?
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!
$\endgroup$ 45 Answers
$\begingroup$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.
$\endgroup$ 6 $\begingroup$Check that the sequences with only $0's$ and $1's$ have dist $1$, and they are uncountable. Can you now finish the problem?
$\endgroup$ 1 $\begingroup$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 $\overline{ A}$.
$\endgroup$ 3 $\begingroup$The set of the sequences with rational coefficients is, indeed, dense in $l^\infty$. However, notice that this set is not countable (I have also thought of it). One can define a surjection from the set of sequences with rational coefficients to the set of parts of natural numbers, which is not countable. So this set is also not countable.
However, it is possible to define the set $Q_n$ as the sequences of rational coefficients until the $n$-th term, and $0$ after the $n-th$ term. The set $$Q = \bigcup_{n=1}^{\infty} Q_n$$ is countable and dense in $l^p$, for $1 \leq p < \infty$, but it does not contain the sequence, for instance, $$(1, 1, 1, 1, 1, ...) \in l^\infty.$$
$\endgroup$ 1 $\begingroup$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.
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