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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?

MAS
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    It don't know about separability but the fact that it is closed is immediate from the definition. – Kavi Rama Murthy Jul 04 '19 at 09:12
  • @KaviRamaMurthy Sir, in the above paper Lorentz claimed that $\mathcal F$ is non-separable. I'm trying to understand it. Would you like to explain about the closeness of $\mathcal F$? – MAS Jul 04 '19 at 09:25
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    I found a proof here, but I seem to be missing something since the author claims on page 79 that the set of all sequences of $0$s and $1$s is contained in $\mathcal{F}$. Which is clearly false. – mechanodroid Jul 04 '19 at 09:52

3 Answers3

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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.

Rhys Steele
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  • From here how can we conclude that $\mathcal F$ is not separable? I think that the closedness of $X$ is necessary. Is it not? Bcz, a topological space is separable if and only if it does not contain an uncountable closed discrete subset. – MAS Jul 05 '19 at 10:57
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    I assume that by $X$ you mean the thing I call $S$ that another answer calls $X$. Well, using your characterisation, note that any sequence in $S$ that converges in $\mathcal{F}$ is Cauchy and in particular eventually the terms are at most distance $\frac{1}{2}$ from eachother. Since distinct terms in $S$ are distance $1$ apart this tells us that the sequence is eventually constant and so its limit lies in $S$. Hence $S$ is closed. – Rhys Steele Jul 05 '19 at 11:06
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    This isn't the usual proof for this setting though. The usual proof says that if ${e_i: i \in \mathbb{N}}$ is a countable dense subset of $\mathcal{F}$ then for every $s \in S$ there is an $e_{i(s)}$ such that $|s - e_{i(s)}| < \frac{1}{2}$. By the triangle inequality, you get that if $s \neq s'$ then $e_{i(s)} \neq e_{i(s')}$ since otherwise $1 = |s-s'| \leq |e_{i(s)} - s| + |s' - e_{i(s)}| < 1$. Hence $s \mapsto e_{i(s)}$ is an injection from $S$ to a countable set. But $S$ is uncountable so this is a contradiction. – Rhys Steele Jul 05 '19 at 11:10
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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 $1/2$.

Jonathan Hole
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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.

mechanodroid
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  • I think in the book of F. Basar(page 78-79), there is some printing mistakes in constructing the uncountable set $X\subset\omega$. Now I've understood through your proper explanation. – MAS Jul 04 '19 at 20:23
  • From your answer it is clear that $X$ is an uncountable discrete subset of $\mathcal F$. But how does it conclude the non-separability of $\mathcal F$? – MAS Jul 05 '19 at 10:42
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    @BijanDatta Let $S$ be a dense subset of $\mathcal{F}$. For every $x \in X$ by density there exists $s_x \in S$ such that $s_x \in B(x, \frac12)$. For $x \ne y$ we have $s_x \ne s_y$ because $ B(x, \frac12)$ and $ B(y, \frac12)$ are disjoint since $|x-y| = 1$. Hence $(s_x)_{x \in X}$ is an uncountable collection of elements in $S$ which makes $S$ uncountable. We conclude that $\mathcal{F}$ cannot be separable. – mechanodroid Jul 05 '19 at 11:48
  • Now it is clear to me that if $S$ be any dense subset of $\mathcal F$, then the uncountability and discreteness of $X$ implies uncountability of $S$. Hence $\mathcal F$ is non-separable. – MAS Jul 05 '19 at 13:51
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    @BijanDatta For a metric space it is equivalent that it is separable and that it has no uncountable discrete subset (and that it has a countable base and many other properties) – Henno Brandsma Jul 09 '19 at 14:23
  • @mechanodroid I need to know that how $x_{n}+\cdots + x_{n+p-1} \le \sqrt{p}$ happens for this sequence $x$? – MAS Aug 07 '19 at 19:24
  • @BijanDatta Actually, it should be $x_{n}+\cdots + x_{n+p-1} \le \sqrt{p}+1$. Indeed, $x_k$ can be nonzero only when $k$ is a perfect square. We have to bound the number of perfect squares in the interval $n, n+1, \ldots, n+p-1$. There are $$\lfloor\sqrt{n+p-1}\rfloor-\lfloor\sqrt{n}\rfloor +\begin{cases} 1, &\text{if $n$ is a perfect square}, \ 0, &\text{otherwise}\end{cases}$$ perfect squares in this range. This can be bounded by $$\sqrt{n+p-1}-\sqrt{n}+1 \le\frac{(n+p-1)-n}{\sqrt{n+p-1}+\sqrt{n}}+1 \ge \frac{p-1}{\sqrt{p}}+1 \le \sqrt{p}+1$$ – mechanodroid Aug 14 '19 at 14:25