The set of all almost convergent real sequences is closed and non-separable in $l_\infty$

Question

Let $l_\infty$ be the normed linear space of all bounded real sequences with the well known sup-norm and $\mathcal F$ be the collection of all almost convergent sequences of real numbers. Then, $\mathcal F$ be a subspace of the topological space $l_\infty$.

Recently, I have been reading this paper by Lorentz (1941). In this paper the author says about some topological properties of $\mathcal F$ (viz. $\mathcal F$ is closed, non-separable)

My Qns : How can I show that $\mathcal F$ is closed and non-separable?

Answer 1

Basically @Rhys Steele's answer but I got it from Summability Theory and Its Applications by Feyzi Basar, linked in the comments:

We have $$\mathcal{F} = \{x \in \ell^\infty : \phi(x) = \psi(x) \text{ for any two Banach limits } \phi, \psi\}$$

If $(x_n)_n$ is a sequence of elements in $\mathcal{F}$ such that $x_n \to x \in \ell^\infty$, then for any two Banach limits $\phi, \psi$ we have

$$\phi(x) = \phi(\lim_{n\to\infty} x_n) = \lim_{n\to\infty}\phi(x_n) = \lim_{n\to\infty}\psi(x_n) = \psi(\lim_{n\to\infty} x_n) = \psi(x)$$ so $x \in \mathcal{F}$. We conclde that $\mathcal{F}$ is closed.

For non-separability consider the set $$X = \left\{x = (x_n)_n\in \ell^\infty : x_n = \begin{cases} 1, &\text{if }n=k^2, k\in S \\ 0, &\text{otherwise}\end{cases}, S \subseteq \mathbb{N} \right\}$$

For each $x \in X$ we have that $x$ has zeroes everywhere except possibly on positions which are perfect squares. Therefore

$$\frac{x_{n}+\cdots + x_{n+p-1}}{p} \le \frac{\sqrt{p}+1}{p} \xrightarrow{p\to\infty} 0$$ uniformly in $n \in \mathbb{N}$ so $x \in \mathcal{F}$ (with every Banach limit being $0$).

We conclude that $X \subseteq \mathcal{F}$ is an uncountable set such that $\|x-y\| = 1$ for all $x,y \in X$, $x \ne y$ so $\mathcal{F}$ is uncountable.

Answer 2

Closedness is straightforward. If $x^{(n)}$ is a sequence in $\mathcal{F}$ that converges to $x$ in $\ell^\infty$ then for any Banach limit $L$, we have that $L(x^{(n)}) \to Lx$ by continuity of $L$. If $L'$ is another Banach limit then $$L'x = \lim_n L'(x^{(n)}) = \lim_n L(x^{(n)}) = Lx$$ so $x \in \mathcal{F}$ and hence $\mathcal{F}$ is closed.

Lorentz tells you how to see this space is non-separable. I'll add some details. He considers the set of sequences $$S := \{x \in \ell^\infty : x_n = 0 \text{ when } n \text{ is not a square and } x_{k^2} \in \{0,1\} \text{ for all } k\}$$ Clearly $|S| = |\{0,1\}^\mathbb{N}| = \mathfrak{c}$, i.e. $S$ is uncountable. We then have that for $x \in S$ and $n \in \mathbb{N}$, $$\frac1p \sum_{i=0}^{p-1} x_{n+i} \to 0$$ as $p \to \infty$ uniformly in $n$, since the left hand side behaves like $\frac{1}{p}$ (After the first contribution to the sum, we have to go at least $k^2$ terms to get a contribution of $1$ to the sum, but then the factor out front has shrunk like $\frac{1}{k^2}$). Hence, by Theorem $1$ of the paper, $S \subseteq \mathcal{F}$. But if $x,y \in S$ are distinct then $\|x-y\|_{\ell^\infty} = 1$ so $\mathcal{F}$ has an uncountable subset where the terms are pairwise distance $1$ apart. This cannot happen in a separable space so $\mathcal{F}$ is not separable.

Answer 3

Closedness is pretty immediate. To show non-seperability it will suffice to find an uncountable collection of almost convergent sequences such that any two of these sequences lie a distance of 1 apart. Indeed $\{\mathcal{X}_S : S\subset \mathbb{N}, S \text{ contains exactly one of } 2n \text{ and } 2n-1 \text{ for each } n\in\mathbb{N}\}$ is such a collection because each of these sequence "converges in average" to $1/2$.