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

Note that from your definition of the limit point of a sequence it follows that each neighborhood of a limit point contains infinitely many terms of the sequence. You can use this statement for your proof as follows:

Suppose for the sake of contradiction that your cauchy sequence $(x_n)$ has two limit points $x \neq y$. Then $\varepsilon := d(x,y) > 0$, by the definiteness of the metric. Since $(x_n)$ is Cauchy, there exists $N \in \mathbb{N}$ such that for all $n, m \geq N$ you have $$d(x_n, x_m) < \frac{\varepsilon}{ 4}.$$ Furthermore, since $x$ is a limit point you find $x_n \in B_{\frac{\varepsilon}{4}}(x)$ with $n \geq N$. But then, for all $m \geq N$ you have
$$
d(x_m, x) \leq d(x_m, x_n) + d(x_n, x) < \frac{\varepsilon}{2}.
$$
So, for all $m \geq N$ this gives you $x_m \in B_{\frac{\varepsilon}{2}}(x)$ which contradicts the assumption that $y$ is another limit point.

## Best Answer

A Cauchy sequence in a metric space $(X,d)$ is not necessarly convergent. for example, consider the metric space $(\mathbf{Q},d)$ where d is the usual absolute value. And consider the sequence $x_{n}$, $x_{1}=1$, $x_{n+1}=\frac{1}{2}(x_{n}+\frac{2}{x_{n}})$. One can check that $x_{n}$ is a Cauchy sequence in $\mathbf{Q}$ but has no limit in $\mathbf{Q}$.