TITLE: Closed subspace of $l^\infty$
QUESTION [12 upvotes]: I've got here this exercise that says: "Show that $c$ is a closed subspace of $l^{\infty}$" (with $c$ I mean the sequences of $l^{\infty}$ that converge in $l^{\infty}$, with respect to the norm of $l^{\infty}$). I've done it, but I cannot say if it is correct.
In order to show that $c$ is a closed subspace of $l^{\infty}$, I have to prove that any convergent sequence $\{c_n\}_n$ of elements of $c$ converges to $x\in c$. I know that, since $l^{\infty}$ is complete, $\{c_n\}$ converges to $x\in l^{\infty}$, so it is enough prove that $x\in c$.
Since $\{c_n\}$ converges to $x\in l^{\infty}$, we have that $||c_n-x||_\infty\to 0$ for $n\to \infty$, i.e. $\sup_{j\in \mathbb{N}}|c_n-x|\to 0$ (here $j$ runs over the elements of the sequence $c_n-x$), i.e. for every $\epsilon>0$ there exists $N>0$ such that $\sup_{j\in \mathbb{N}}|c_n-x|<\epsilon$ for $n>N$. 
Now, since $c_n\in c$, we have that $c_n\to \xi\in l^{\infty}$ for $n\to \infty$, and so for every $\epsilon>0$  
\begin{equation*}
\sup_{j\in \mathbb{N}}|\lim_{n\to \infty}(c_n-x)|=\sup_{j\in \mathbb{N}}|\xi-x|<\epsilon, 
\end{equation*}
which means that $x\in c$.
What do you think? Is there anybody that could suggest me a different argument?

REPLY [6 votes]: I think your argument has the right ideas, but it still needs a little touch since your $\xi$ depends on $n$. 
It is probably easier to just show that $x$ is Cauchy: for $\varepsilon>0$, choose $k$ with $\|x-c_k\|_\infty<\varepsilon$, and $n_0$ such that $\|c_k(m)-c_k(n)|<\varepsilon$ for all $m,n>n_0$. Then
$$
|x(m)-x(n)|\leq |x(m)-c_k(m)|+|c_k(m)-c_k(n)|+|c_k(n)-x(n)|\\\leq2\|x-c_k\|_\infty+|c_k(m)-c_k(n)| 
\leq3\varepsilon.
$$