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?
Best Answer
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.