$\ell^\infty(\mathbb{N})$ is not separable

Question

Show, that $\ell^\infty(\mathbb{N})$ is not separable. Use, that $\mathcal{P}(\mathbb{N})$ is uncountable and observe the set $\{1_A\colon A\in\mathcal{P}(\mathbb{N})\}\subseteq\ell^\infty(\mathbb{N})$.

Suppose $\ell^\infty(\mathbb{N})$ is separable. Then it exists a countable, dense set $D=\{d_n\colon n\in\mathbb{N}\}\subseteq\ell^\infty(\mathbb{N})$.

Since $D$ is dense in $\ell^\infty(\mathbb{N})$ exists for every $a_k\in\ell^\infty(\mathbb{N})$ and $\varepsilon>0$ a sequence $(d_k^{(n)})\subseteq D$ with $\|d_k^{(n)}-a_k\|_{\infty}=\|d_k(n)-a_k(n)\|_{\infty}<\varepsilon$

Now I want to construct an $(a_k)$ such that it can not be the limit of any $d_k\in D$.

I am trying to follow the hint in the task:

$a_k=\begin{cases} 0,~~ \text{if}~~ |d_k(k)|>1\\ 1,~~ \text{else}\end{cases}$

Now we have:

$\vert|d_k(n)-a_k(n)\|_\infty=\sup_{n\in\mathbb{N}} |d_k(n)-a_k(n)|\geq 1$ for all $n\in\mathbb{N}$

This is a contradiction. Hence $\ell^\infty(\mathbb{N})$ is separable.

Is this correct? Thanks in advance for your help.

Answer 1

I think your idea is basically okay, but your notation is a little bit wrong. You should consistently write $a(k)$ and $d_n(k)$ to mean the $k$'th entries of the sequences $a$ and $d_n$ (sometimes you are writing things like $a_k$, or $d_k^{(k)}$, the switching notation is a little unclear).

Since you are given a countable set $\{d_n \mid n \in \mathbb{N}\}$, you can define $a$ by $a(k) = 0$ or $2$, if $|d_k(k)|$ is $>1$ or $<1$. (This is changing $a(k)=1$ to $2$. The point is that this way, $|a(k)-d_k(k)| > 1$.)

Now you have $\|a-d_k\| > 1$ for all $k$, so $d_k \not\to a$. Your idea works! Just use notation more consistently (and change $1$ to $2$).

Having said that, @Ian's comments are right too. You can use the hint more directly. You can show that if $A \neq B$, then $\|1_A-1_B\| = 1$. I won't say too much more, but knowing $\|1_A-1_B\|=1$ tells you something about the balls centered at the $1_A$'s, which you can use to show $\ell^{\infty}$ is nonseparable. I hope that helps!

Answer 2

Here are two ways to use the hint:

Way 1: Use the subspace topology.

Fact 1: A discrete topological space $X$ is separable if and only if it is countable. (This is easy enough that I don't think you should need help with it.)

Fact 2: Every subspace $A$ of a separable metric space $X$ is separable in its subspace topology. This is less obvious, so I'll sketch the proof. Enumerate your dense subset of $X$ as $d_n$, take a sequence $r_k$ of positive real numbers going to zero, and draw one point from $A \cap B(d_n,r_k)$ whenever this is possible. The result is a countable subset of $A$ which is dense in the subspace topology of $A$. (Note that it won't be dense in $X$ if $A$ wasn't dense in $X$.)

The overall logic of Way 1: if $X$ contains an uncountable discrete subspace, then it cannot be separable, because if it were then the uncountable discrete subspace would also be separable.

Way 2: Use the pigeonhole principle, in the form "if $|A|<|B|$ then there is no surjection from $A$ to $B$".

We have the open balls $B(1_A,1/2)$ ranging over the subsets of $\mathbb{N}$. They are disjoint by the hint. Thus given any countable subset $D$ of $\ell^\infty$, we can excise the countably many balls that intersect $D$ from the union of all these balls, obtaining a nonempty open subset of $\ell^\infty$ which does not intersect $D$.