Space of bounded sequences in Banach space is complete Let $(X, |\cdot|)$ be a Banach Space and 
$$X_b := \{ (x_n)_{n\in \mathbb{N}} \subset X: \| (x_n)_{n\in\mathbb N} \|_{X_b}< \infty \}$$
 with 
$$\|(x_n)_{n\in\mathbb N}\|_{X_b}:= \sup\limits_{n \in \mathbb{N}} |x_n|.$$
Why is $(X_b,\|\cdot\|_{X_b})$ complete?
 A: Suppose $(z_n)_{n\in\mathbb N}$ is a Cauchy sequence in $X_b$, with $z_n=(x_{n,m})_{m\in\mathbb N}$ for each $n$.  Fix $\varepsilon>0$.  Then there is some $N\in\mathbb N$ such that 
$$\|z_{n_1}-z_{n_2}\|_{X_b}<\varepsilon$$
for all $n_1,n_2\geq N$.  But then for each $m\in\mathbb N$ we have 
$$|x_{n_1,m}-x_{n_2,m}|\leq\|z_{n_1}-z_{n_2}\|_{X_b}<\varepsilon,$$
This means that the sequence $(x_{n,m})_{n\in\mathbb N}$ is Cauchy in $X$, and thus convergent to some $x_m\in X$.  This gives us a sequence $z=(x_m)_{m\in\mathbb N}$ in $X$.  Now we have to show two things:
$(1)$ $z\in X_b$, and
$(2)$ $\|z_n-z\|_{X_b}\to0$ as $n\to\infty$.
The first isn't too difficult.  Given $m\in\mathbb N$, there is some $n\in\mathbb N$ such that $|x_m-x_{n,m}|<\varepsilon$, and thus 
$$|x_m|<|x_{n,m}|+\varepsilon\leq\|z_n\|_{X_b}+\varepsilon\leq\sup_{n\in\mathbb N}\|z_n\|_{X_b}+\varepsilon,$$
and we know $\sup_{n\in\mathbb N}\|z_n\|_{X_b}<\infty$ since $(z_n)_{n\in\mathbb N}$ is Cauchy.  Since $m\in\mathbb N$ was arbitrary, we have 
$$\sup_{m\in\mathbb M}|x_m|\leq\sup_{n\in\mathbb N}\|z_n\|_{X_b}+\varepsilon<\infty,$$
thus $z\in X_b$, so $(1)$ is shown.
Now to show $(2)$, since $(z_n)_{n\in\mathbb N}$ is Cauchy, there is some $N'\in\mathbb N$ such that $\|z_{n_1}-z_{n_2}\|_{X_b}<\varepsilon/2$ whenever $n_1,n_2\geq N'$.  Now let $n\geq N'$ be given.  Then for any $m\in\mathbb N$, there is some $n'\geq N'$ such that $|x_m-x_{n',m}|<\varepsilon/2$.  Thus we have 
$$|x_m-x_{n,m}|\leq|x_m-x_{n',m}|+|x_{n',m}-x_{n,m}|<\varepsilon/2+\varepsilon/2=\varepsilon.$$
Now taking the supremum over $m$, we have 
$$\|z-z_n\|_{X_b}\leq\varepsilon,$$
and thus $(2)$ is shown.
