47
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ math.stackexchange.com/questions/97648/… $\endgroup$ Commented Feb 2, 2014 at 7:19
  • 3
    $\begingroup$ 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. $\endgroup$
    – user67803
    Commented Feb 2, 2014 at 7:25
  • 3
    $\begingroup$ 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. $\endgroup$
    – user67803
    Commented Feb 2, 2014 at 7:36
  • $\begingroup$ @user67803 : notice the countably-infinite product of factors of the set $\{0,1,2,3,4,..,9\}$ is the set of all possible decimal expansions $.a_1a_2...a_n.....$ which is of the same cardinality of the Real numbers. $\endgroup$
    – MSIS
    Commented Dec 22, 2021 at 23:01

5 Answers 5

81
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ Could you explain why these balls are disjoint, I don't see why they have to be (it's clear they're open) $\endgroup$
    – PSub
    Commented Oct 6, 2017 at 2:10
  • 6
    $\begingroup$ @PSub: the distance of centers of different balls is $1$ but the radius of $B$ is $1/2$. $\endgroup$
    – user9464
    Commented Oct 6, 2017 at 2:14
  • $\begingroup$ Thanks @Jack, yeah I figured that out eventually, it wasn't evident to begin with. $\endgroup$
    – PSub
    Commented Oct 6, 2017 at 3:25
  • 1
    $\begingroup$ Should $(e_I)_i$ be defined by $0$ if $i\not\in I$ instead of $1\not\in I$? $\endgroup$ Commented Apr 17, 2018 at 17:06
  • 1
    $\begingroup$ @mabavilj The set of such sequences is in bijection with the power set of $\Bbb{N}$, hence it is uncountable. $\endgroup$ Commented Oct 23, 2019 at 11:19
49
$\begingroup$

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
  • 17
    $\begingroup$ 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. $\endgroup$
    – triple_sec
    Commented Feb 2, 2014 at 8:26
16
$\begingroup$

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
  • 1
    $\begingroup$ 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. $\endgroup$
    – Derek Orr
    Commented Sep 10, 2017 at 22:49
  • 3
    $\begingroup$ 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 $(a^k)=(k,k,k,...)$. $\endgroup$ Commented Sep 12, 2017 at 13:47
  • $\begingroup$ @alexb What is the idea to define $b_k$? Is it started from $|b_k−a^k_k|≥1$ for every $k$? So if we define $b_k=0$, then we have $|0−a^k_k|=|a^k_k|≥1$. And for $|a^k_k|<1$, we can define $b_k=a^k_k+1$ to make sure that $|b_k|=|a^k_k+1|≤|a^k_k|+1<1+1=2$ which implies $\sup_{k \in \mathbb{N}} |b_k|<\infty$. $\endgroup$
    – user136524
    Commented Jan 5, 2022 at 7:15
7
$\begingroup$

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
  • $\begingroup$ I don't see why it should have to contain the sequence you mention. It is just a subset of it. $\endgroup$
    – MSIS
    Commented Dec 22, 2021 at 23:41
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @Didier: Thank you, corrected. $\endgroup$
    – MSIS
    Commented Dec 22, 2021 at 23:38

You must log in to answer this question.