# Subspace of $\ell_\infty$ that is not separable

I need to prove that

$$L = \left\{ (x_i)\in \ell_\infty : \lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n x_i = 0\right\}$$

is a subspace in $l_\infty$ (done that), is closed in $l_\infty$ (done that), and not separable in $l_\infty$.

How would I go about proving that it is not separable?

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Prove that $L$ has an uncountable discrete subset – Ivo Mar 5 '13 at 17:25

Fix a set $T\subseteq N$ that is infinite but so thin that your $L$ contains the characteristic function of $T$, i.e., the sequence that has $1$ in position $i$ for all $i\in T$ and $0$ in all other positions. For example, $T$ could be the set of all powers of $2$. Consider the characteristic functions $\chi_A$ of arbitrary subsets $A$ of $T$. There are continuum many such subsets $A$ and therefore continuum many such $\chi_A$'s. All these $\chi_A$'s are in $L$ because of our choice of $T$. The $\ell_\infty$ distance between any two distinct $\chi_A$'s is $1$, so no point is within $1/2$ of more than one of them. Therefore, any dense set (or any set that even gets within $1/2$ of each of our $\chi_A$'s) must have the cardinality of the continuum. (Matt N. had a similar answer, but as far as I can see he overestimated the number of $y_i$'s in his answer.)
What is $N$? ${}$ – Rudy the Reindeer Mar 5 '13 at 17:37
$N=\mathbb N=\omega=$ the set of natural numbers. – Andreas Blass Mar 8 '13 at 22:11