TITLE: How to show that a Cauchy sequence of sequences converges
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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).