[Math] How to show that a Cauchy sequence of sequences converges

Question

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?

Answer 1

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 $d(x,x_n) \le \epsilon$.