Convergent sequences are closed subspace of $l^\infty$ 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 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 { $y$ Cauchy } $\Rightarrow$ {$y$ 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.
Consider $l^{\infty}\setminus C$
Let $y_{n} \in l^{\infty}\setminus C\,,$ then $\forall y \in\mathbb{R}, \exists \varepsilon_{y} &gt; 0$ such that $\|y-y_{n}\|_{\infty} &gt; 2\varepsilon_{y}$
Now consider $\mathbb{B}(y_{n},\varepsilon_{y})$
Let $z_{n} \in \mathbb{B}(y_{n},\varepsilon_{y})\Longrightarrow \|y_{n}-z_{n}\|_{\infty}&lt;\varepsilon_{y}$
So $\forall y \in \mathbb{R}$;
$\|y-z_{n}\|_{\infty} = \|y-y_{n}-(z_{n}-y_{n})\|_{\infty} \geq \|y-y_{n}\|_{\infty} - \|z-y_{n}\|_{\infty} &gt; 2\varepsilon_{y} - \varepsilon_{y} = \varepsilon_{y} &gt; 0$
$\Longrightarrow \mathbb{B}(y_{n},\varepsilon_{y}) \subseteq l^{\infty}\setminus C\Longrightarrow l^{\infty}\setminus C$ is open $\Longrightarrow C$ is closed.
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 $\ell^{\infty}$.