Show that $c$ is closed in $l^{\infty}$

Question

Let $$c=\{ (a_i)_{i \in \mathbb{N}} \ ; \ \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N} \ , \ \mbox{exist} \ \displaystyle \lim_{i \to \infty}(a_i)\}$$ $$l^{\infty}=\{ (a_i)_{i \in \mathbb{N}} \ ; \ a_i \in \mathbb{R}\ ,\ \forall i \in \mathbb{N}; \ \displaystyle \sup_{i \in \mathbb{N}}|a_i|< \infty \}.$$ Show that $c$ is closed in $l^{\infty}$.

i am trying... taking a sequence of sequence in $c$ but i can't...i show that $c_0$ is closed in $l^{\infty}$.

Answer 1

Let $f : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \limsup\limits_{n \to + \infty} x_n \end{array} \right.$ and $g : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \liminf\limits_{n \to + \infty} x_n \end{array} \right.$.

Show that $f$ and $g$ are continuous, and deduce that $c= \{ x \in \ell^{\infty} \mid f(x)=g(x) \}$ is closed in $\ell^{\infty}$.

Answer 2

Hint: a subset $A$ of a metric space $X$ is closed if and only if for any sequence $a_n\in A$ converging to an $x\in X$ you have $x\in A$.

Answer 3

Hint 1: Convergence in $\ell_\infty$ and as a consequence in $c$ is a uniform convergence on $\mathbb{N}$

Hint 2: If $\{a^{(n)}:n\in\mathbb{N}\}\subset c$ is a uniformly convergent sequence, then $$ \lim\limits_{i\to\infty}\lim\limits_{n\to\infty}a_i^{(n)}= \lim\limits_{n\to\infty}\lim\limits_{i\to\infty}a_i^{(n)} $$

Answer 4

Let $x=(\xi_1,\xi_2,\ldots)\in \bar{c}$ and let $(x_n)$ be a sequence in $c$ such that $lim \ x_n = x$ and for each $n$ we write $x_n$ as $(\xi_1^n,\xi_2^n, \dots )$

Part 1)

By the definition of convergence and using the norm of $l^{\infty}$, we have that $\forall \epsilon >0 \ \exists N_0 \in \mathbb{N}$ such that $n \geq N_0 \implies \|x_n - x \| = sup_{j\leq 1} \ |\ \xi^n_j - \xi_j| < \epsilon/3$

Also for each fixed $j$ we have $| \xi^n_j - \xi_j| < \epsilon/3$.

Part 2)

Let $n=N_0$, $x_N \in c$ and in particular $x_N$ is a Cauchy sequence. So $\forall \epsilon > 0 \ \exists N_1 \in \mathbb{N}$ such that $j,k\geq N_1 \implies |\xi^N_j - \xi^N_k| < \epsilon/3$

Part 3)

Given $\epsilon >0$, taking $j,k \geq max{\{N_1, N_0}\}$

$|\xi_j - \xi_k| \leq |\xi_j - \xi^N_j| + |\xi^N_j - \xi^N_k| + |\xi^N_k - \xi_k| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$

This proves $x$ is a Cauchy sequence.

Part 4)

$K =$ $\mathbb{R}$ or $\mathbb{C}$ is complete AND $x \in K^\infty$ AND $x$ is a Cauchy sequence, therefore x converges, i.e., $x \in c$.

Equivalently, this proves that $c$ is a complete vector subspace of $l^\infty$.