Closed subspace of $l^\infty$ 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?
 A: 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
$$
|x(m)-x(n)|\leq |x(m)-c_k(m)|+|c_k(m)-c_k(n)|+|c_k(n)-x(n)|\\\leq2\|x-c_k\|_\infty+|c_k(m)-c_k(n)| 
\leq3\varepsilon.
$$
A: Careful book-keeping is needed for a proof like this. Denote by $c_n = (c_n^{(j)})_{j\in\mathbb{N}}$ a generic term of a sequence of elements of $l^\infty$. Define $c:=\{(x^{(j)})_{j\in\mathbb{N}}\in l^\infty ~|~ \exists \lim\limits_{j\rightarrow\infty} x^{(j)} \in \mathbb{C} \}$, a subspace of $l^\infty$. Note that the elements of $c$ are precisely the convergent sequences in $\mathbb{C}$ (since these are automatically bounded). To avoid confusion, I will refer only to convergence in $\mathbb{C}$ using the notation $\lim\limits_{j\rightarrow\infty}$.
Now suppose the sequence $(x_n)_{n\in\mathbb{N}}$ of elements of $c$ has a sup-norm limit $x$. By completeness, as you said, $x = (x^{(j)})_{j\in\mathbb{N}} \in l^\infty$, i.e., it's a bounded sequence of complex numbers, and by the definition of $l^\infty$-convergence we have that $\|x_n-x\|_\infty\rightarrow 0$ as $n\rightarrow \infty$.
To prove that $x \in c$ we need to show that there is $\xi \in \mathbb{C}$ which is the limit of $(x^{(j)})_{j\in\mathbb{N}}$, i.e., $\lim\limits_{j\rightarrow\infty} x^{(j)} = \xi \in \mathbb{C}$. The obvious candidate is the limit (if it exists!) of the sequence $(\xi_n)_{n\in\mathbb{N}}$, where $\xi_n:=\lim\limits_{j\rightarrow\infty} x_n^{(j)}$ exists in $\mathbb{C}$ for each $n\in\mathbb{N}$ since each $x_n$ is in $c$.
Now, $|\xi_n - \xi_m| \leq |\xi_n - x_n^{(j)}| + \|x_n- x_m\|_\infty + |x_m^{(j)} - \xi_m| \rightarrow 0$, as $m,n\rightarrow \infty$ (since the LHS is independent of the $j$ in the RHS, so we can take $j\rightarrow \infty$ as well). Using the completeness of $\mathbb{C}$, there is a limit $\xi = \lim\limits_{n\rightarrow\infty} \xi_n \in \mathbb{C}$.
Then, $|x^{(j)}-\xi| \leq \|x - x_n\|_\infty + |x_n^{(j)} - \xi_n| + |\xi_n - \xi|\rightarrow 0$ as $j\rightarrow \infty$ since, similarly to earlier, the LHS is independent of $n$ in the RHS. The proof is complete.
A: For any $\epsilon$ it is true that $\|x-c_n\|\leq \epsilon$ and $\|c_n\|\leq M$ for some $M>0$(since $c_n\in\ell_{\infty}$). Thus you have $\|x\|_\infty\leq \epsilon+M$ (Choose $\epsilon=1$).
