I have stumbled upon a particular proof of a fact that $\ell^\infty$ is not separable which I can't quite grasp:
- For $M\subset \mathbb N$ define $\chi_M(n)=1$ if $n\in M$ and $0$ otherwise. $\chi_M \in \ell^\infty$ since it's bounded sequence.
- Let $A$ be a countable subset of $\ell^\infty$. For every $x \in A$ the set $\{ y \in \ell^\infty\ |\ \Vert x-y \Vert \leq 1/4\}$ contains at most one element of uncountable set $\{ \chi_M| M \subseteq \mathbb N\}$.
- $\therefore A $ is not dense.
The conclusion from 2. to 3. isn't obvious to me.
For $A$ to be not dense would mean an existence of open non empty set that does not contain any element of $A$. But part 2. does not provide explicitly any of it.
Question: What are logical steps from 2. to 3.?
Best Answer
For the set $A$, to be dense, its closure has to be the entire space $\ell^\infty$.
For any element in $\ell^{\infty}\setminus A$ to be in the closure of $A$, it must at least fall within distance $\frac{1}{4}$ of some element of $A$. (This is necessary, though CLEARLY not sufficient.)
The claim (2) really just says that the set of points of $\ell^{\infty}$ that lie within $\frac{1}{4}$ of ANY element of $A$ is at most countable; therefore the closure (which must be a subset of this) is also at most countable.
Since $\ell^{\infty}$ is uncountable, the closure cannot contain every element of $\ell^\infty$.