Convergent sequences are closed subspace of $l^\infty$

Question

Here is the exercise:

Let's consider the Banach space $l^{\infty}$ with the sup norm.

We consider the following subspace of $l^{\infty}$ :

$C = ${ $x \in l^{\infty} : \lim_{n\to \infty} x_n\quad\text{ exists }$}.

We have to show that $C$ is a closed subspace of $l^{\infty}$ and is thus also a Banach space with respect to the sup norm.

Let $y$ denote $\lim_{n\to \infty}$ $x_n$.

The key there is to show that every convergent sequence in $C$, say every sequence in $C$ converges within $C$, hence we have to show that $y$ belongs to $C$. I've successfully shown that $y$ is a Cauchy sequence in $l^{\infty}$, but I don't know what to do with that. When I'm looking at the correction, it says that $\{ y\text{ Cauchy}\} \Rightarrow \{y\text{ convergent}\} \Rightarrow \{y \in C\}$.

That's where I'm lost. I don't understand why we can deduce $y$ is convergent by having shown $y$ is Cauchy. Also, I don't understand why we have to show that $y$ is Cauchy.

I hope someone can explain me those two problems I have.

I do understand though that $C$ being closed implies that $C$ is a Banach space.

Answer 1

Let $(x_n)_n$ be a Cauchy sequence in $C$ converging to $y=(y_j)_j\in l_{\infty}.$ Let $x_n=(x_{n,j})_j.$ For each $j$ we have $\lim_{n\to \infty}x_{n,j}=y_j . $

By contradiction: Suppose $y$ is not a convergent sequence. Then for some $r>0$, we have $\forall n\in N\;(\exists j,k\;(n\leq j<k\land |y_j-y_k|>r).$

But let $\|x_{n}-y\|<r/4$ and let $j_0$ satisfy $(j\geq j_0\implies |x_{n,j}-L(n)|<r/4),$ where $L(n)=\lim_{j\to \infty}x_{n,j}.$

Then there exists $j\geq j_0$ and $k>j$ such that $|y_j-y_k|>r.$

This yields $$|y_j-L(n)|\leq |y_j-x_{n,j}|+|x_{n,j}-L(n)| <\|y-x_n\|+r/4<r/2$$ and $$|y_k-L(n)|\leq |y_k-x_{n.k}|+|x_{n,k}-L(n)|< \|y-x_n|+r/4<2.$$ But then we have $|y_j-L(n)|<r/2$ and $|y_k-L(n)|<r/2,$ which implies $|y_j-y_k|<r,$ a contradiction.

Answer 2

Consider $l^{\infty}\setminus C$
Let $y_{n} \in l^{\infty}\setminus C\,,$ then $\forall y \in\mathbb{R}, \exists \varepsilon_{y} > 0$ such that $\|y-y_{n}\|_{\infty} > 2\varepsilon_{y}$
Now consider $\mathbb{B}(y_{n},\varepsilon_{y})$
Let $z_{n} \in \mathbb{B}(y_{n},\varepsilon_{y})\Longrightarrow \|y_{n}-z_{n}\|_{\infty}<\varepsilon_{y}$
So $\forall y \in \mathbb{R}$;
$\|y-z_{n}\|_{\infty} = \|y-y_{n}-(z_{n}-y_{n})\|_{\infty} \geq \|y-y_{n}\|_{\infty} - \|z-y_{n}\|_{\infty} > 2\varepsilon_{y} - \varepsilon_{y} = \varepsilon_{y} > 0$
$\Longrightarrow \mathbb{B}(y_{n},\varepsilon_{y}) \subseteq l^{\infty}\setminus C\Longrightarrow l^{\infty}\setminus C$ is open $\Longrightarrow C$ is closed.

Answer 3

Let $X=\{1,2,3,\cdots\}\cup\{\infty\}$ and define a topology on $X$ that is discrete, but neighborhoods of $\infty$ include all large integers. Then $X$ is compact, and your sequences are the same as the continuous functions $C(X)$. This is the one-point compactification. $C(X)$ is a Banach space, and the elements of $C(X)$ are determined by their values on the integers, with a natural isometric inclusion in $\ell^{\infty}$.