Some space of sequences is complete. Let A be a Banach space and B be the space of all sequences in A that converge to 0, with norm being the maximum element of the sequence. I need to prove that B is complete.
What I've done so far: I've proved that a Cauchy sequence (of sequences) is Cauchy pointwise, therefore converges pointwise because A is Banach. I also proved that this "suspect" (the sequence of pointwise limits) is in B. However, I don't know how to prove convergence with the norm in B.
 A: Let $(x_i)$ be the pointwise limit of $(x_i)_n \in B$, $x_i \in A$. 
Let $\varepsilon > 0$. We know $(x_i)_n$ is Cauchy with respect to $\|(x_i)\| = \sup_i |x_i|$. Hence there is $N$ such that $n,m > N$ implies $ \|(x_i)_m - (x_i)_n\| < \varepsilon$. 
It follows that for $n>N$ $\lim_{m \to \infty} \|(x_i)_m - (x_i)_n\| = \|(x_i) - (x_i)_n\| \le \varepsilon$.
A: I'll treat a more general case. To recover your case, just make $X=\mathbb{N}$, $E=\mathbb{R}$ and $x_0=+\infty$.
The set $B(X,E)$ of bounded $E$-valued functions on some set $X$, where $E$ is a Banach space, is a Banach space when equipped with the sup norm
$$
\|f\|_\infty=\sup_{x\in X}\|f(x)\|_E.
$$ 
This goes exactly like proving that $\mathbb{R}^n$, or even $\ell^\infty(\mathbb{R})$ is complete, using the fact that $\mathbb{R}$ is complete. A uniform Cauchy sequence is pointwise Cauchy, etc...
Now assume $X$ is a topological space. For every limit $\lim_{x_0}$ which makes sense on $X$ (e.g $x_0$ in $X$ metric space, or $x_0=+\infty$ in $\mathbb{R}$ or in $\mathbb{N}$), 
$$
B_{x_0}(X,E):=\{f\in B(X,E)\;;\;\lim_{x_0}f=0\}
$$
is closed in $B(X,E)$, hence complete.
Proof: That's an $\epsilon/3$ proof. Let $f_n$ in $B_{x_0}(X,E)$ tend to $f$ in $B(X,V)$.
Take $\epsilon>0$. 
There exists $N$ such that $\|f_N-f\|_\infty\leq \epsilon/3$.
Since $\lim_{x_0}f_N=0$, there is a neighborhood $U$ of $x_0$ such that
$$
\|f_N(x)-f_N(x_0)\|_E\leq \epsilon/3\qquad\forall x\in U.
$$
Then
$$
\|f(x)-f(x_0)\|_E\leq\|f(x)-f_N(x)\|_E+\|f_N(x)-f_N(x_0)\|_E+ \|f_N(x_0)-f(x_0)\|_E
$$
$$
\leq \frac{\epsilon}{3}+ \frac{\epsilon}{3}+ \frac{\epsilon}{3}=\epsilon
$$
for all $x\in U$.
So $\lim_{x_0}f=0$ and $f$ belongs to $B_{x_0}(X,E)$. Hence $B_{x_0}(X,E)$ is closed.
