# 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 $l^\infty$?

• math.stackexchange.com/questions/97648/… – mathematician Feb 2 '14 at 7:19
• I suppose my error is assuming that the countably infinite cartesian product of countable sets is countable. I got this hunch from Cantor's diagonalization argument for rational numbers. I'm still working on why this is not the case in general. – fierydemon Feb 2 '14 at 7:25
• I just got my fallacy. Cantor's argument for rational numbers only proves $\Bbb{Z}\times\Bbb{Z}$ is countable. This is not an infinite product of countably infinite sets. – fierydemon Feb 2 '14 at 7:36

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

• Some further hint: Use @voldemort's answer to establish the existence of uncountably many pairwise disjoint nonempty open sets, and argue that any dense set must have at least one of its points lie in any nonempty open subset of the space. – triple_sec Feb 2 '14 at 8:26

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.

• Could you explain why these balls are disjoint, I don't see why they have to be (it's clear they're open) – PSub Oct 6 '17 at 2:10
• @PSub: the distance of centers of different balls is $1$ but the radius of $B$ is $1/2$. – Jack Oct 6 '17 at 2:14
• Thanks @Jack, yeah I figured that out eventually, it wasn't evident to begin with. – PSub Oct 6 '17 at 3:25
• Should $(e_I)_i$ be defined by $0$ if $i\not\in I$ instead of $1\not\in I$? – MathStudent1324 Apr 17 '18 at 17:06
• Why is the set of balls uncountable? – mavavilj May 26 '19 at 17:40

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 $\overline{ A}$.

• Why do we need $|a_{k}^{k}| \leq 1$? Can't we just define the sequence $b$ as $b_{k} = a_{k}^{k}+1$? Then $|b_{k}-a_{k}^{k}| = 1$ which is less than or equal to the sup-norm. – Derek Orr Sep 10 '17 at 22:49
• Because then you might have an unbounded sequence if $a^k_k \to \infty$. Each sequence $(a^k)$ is bounded by an $M_k$, but you could have $M_k\to\infty$ as $k\to\infty$. For example, let $(a^k)=(k,k,k,...)$. – AspiringMathematician Sep 12 '17 at 13:47

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, $$(1, 1, 1, 1, 1, ...) \in l^\infty.$$