# $\ell^\infty$ Is Not Separable

## Background

This Wednesday, I read a proof about the non-separability of $\ell^\infty$ spaces. To simplify things, I assume that it’s defined on sequences.

$L^\infty \triangleq \{ (x_n)_{n \in \N} \mid \exists\,M \ge 0 \text{ such that } \forall\,n \in \N, \abs{x_n} \le M \}$

I have written down the uncountable set (see Cantor’s diagonal argument)

$D \triangleq \{ (x_n)_{n \in \N} \mid \forall\,n \in \N, x_n=\pm1 \}$

in my notes. I understand

$B(x,1) \cap B(y,1) = \varnothing \quad \forall \, x,y \in D \text{ with } x \ne y.$

## Problem

Then, my teacher said that the reason for the non-separability of $\ell^\infty$ was like the Pigeon-Hole Principle. I got puzzled when I was revising the proof. In fact, in the above equation, an element in a dense set $C$ can be found in each open ball $B(x,1)$. Since the open balls $B(x,1)$ are disjoint, $C$ has uncountably many elements. Hence, $\ell^\infty$ is non-separable.

One knows that if a space is separable, we can’t insert a non-separable subspace into it. The above property of $\ell^\infty$ serves as a concrete example of this fact.

It’s easy to show that a space having a Schauder basis is separable. Thus, we can conclude that $\ell^\infty$ don’t possess any Schauder basis.

In fact, we have a more direct approach to the absence of Schauder basis in $\ell^\infty$. This will be discussed in my next post.