How to show that a Cauchy sequence of sequences converges Let $c_0$ be the space of real-valued sequences $\{x_n\}$ which converge to
zero, equipped with the metric $d(\{x_n\}, \{y_n\}) = sup_n
|x_n − y_n|$. I want to show that the metric space $(c_0, d)$ is complete, in other words every Cauchy sequence converges. I need to show that every Cauchy sequence has a limit, and that that limit actually belongs to $c_0$. Right now I am having trouble with the first part. If we assume that $\{a_n\}$ in $c_0$ is Cauchy, I know by the definition of the metric $d$ that the real valued sequence along the $k^\text{th}$ term of each term (which itself is a sequence of reals) of $\{a_n\}$ is also Cauchy. In other words, the sequence $\{a_n\}$ converges term-wise (since Cauchy sequences in the reals converge), but how do I use that to show the entire sequence converges? 
 A: There is a notational issue here. You need to show that a Cauchy sequence of elements of $c_0$ converges to an element of $c_0$. It might help to refer to an element of $c_0$ as $x$, and then address a specific element of the sequence as $x(k)$. 
So suppose you have $x_n \in c_0$ such that $x_n$ is Cauchy with the distance
given above.
For each $k$, you have $|x_n(k)-x_m(k)| \le d(x_n,x_m)$, so there is some $x(k)$
such that $x_n(k) \to x(k)$. This is the candidate sequence.
Now you must show that $x \in c_0$ and $d(x,x_n) \to 0$.
Let $\epsilon>0$, choose $N$ such that if $m,n \ge N$ then
$d(x_n,x_m) < { 1\over 3} \epsilon$. Now choose $K$ such that 
if $k \ge K$, then $|x_N(k) | < { 1\over 3} \epsilon$. Then, for $k \ge K$
\begin{eqnarray}
|x(k)| &\le& |x(k)-x_m(k)| + |x_m(k)-x_N(k)| + |x_N(k)| \\
&<& |x(k)-x_m(k)| + {2 \over 3}\epsilon
\end{eqnarray}
Now choose $m \ge N$ (which will, in general, depend on $k$) such that
$|x(k)-x_m(k)| < {1\over 3} \epsilon$, and we see that
$|x(k)| <\epsilon$. Hence $x (k) \to 0$ and so $x \in c_0$.
Showing that $x_n \to x$ is similar:  Let $\epsilon>0$ and choose
$N$ such that if $m,n \ge N$ then
$d(x_n,x_m) < { 1\over 2} \epsilon$. Then, for $m,n \ge N$ we have
\begin{eqnarray}
|x(k)-x_n(k)| &\le& |x(k)-x_m(k)| + |x_m(k)-x_n(k)| \\
&<& |x(k)-x_m(k)| + {1 \over 2} \epsilon
\end{eqnarray}
Now choose $m \ge N$ (which will, in general, depend on $k$) such that
$|x(k)-x_m(k)| < {1 \over 2} \epsilon$, then we see that
$|x(k)-x_n(k)| < \epsilon$ for all $n \ge N$, and so $d(x,x_n) \le \epsilon$.
A: Let $\{x_n\}$ be a Cauchy sequence. Then, $d(x_n,x_m) \to 0$ as $n,m \to \infty$.
Let $c \in \mathbb{N}$. Note that for all $m,n$, $|x_m(c)- x_n(c)| \leq d(x_m-x_n)$.
Hence, $x_m(k)$ is a Cauchy sequence in $\mathbb{R}$, for all fixed $k \in \mathbb{N}$. Moreover, $x_m(k)$ is uniformly convergent, which is to say,  that given $\epsilon > 0$, there is $M$ large enough so that $|x_m(k)-x(k)| < \epsilon$  for $m > M$, for all $k \in \mathbb{N}$.
Now, $\mathbb R$ is complete, so there exists $l_k$ such that $x_m(k) \to l_k$.
Define a new sequence $x(k) = l_k$.
See that $d(x_n , x) = \sup |x(d)-x_n(d)|$. By uniform convergence property, this converges to zero, hence $x_n \to x$.
To see that $x \in c_0$, we want that $x(d) \to 0$ as $d \to \infty$. We know that $x(d) = \lim x_n(d)$, so use this knowledge, along with the fact that $x_n(d) \to 0$  as $n \to \infty$, to show that $x(d) \to 0$. (It's a limit switching argument).
