Show that $(X,d)$ is complete.

Question

Let $(X,d)$ be a metric space where $d$ is the discrete metric. That is: $$\begin{equation} d(x,y) =\left\{ \begin{array}{@} 0,\,\, if \,x = y,\\ 1, \,\,if\, x \neq y. \end{array}\right. \end{equation}$$

I need to show that $(X,d)$ is complete (that every Cauchy-sequence in X converges to some point in X).

What I've tried to do is this:

Pick an arbitrary sequence $(x_n)$ in $X$ and suppose that it is Cauchy. That is, suppose that for each $\epsilon > 0$ there exists an $n_0$ such that $|x_n - x_m|< \epsilon$, whenever $m,n\geq n_0$. The fact that $(x_n)$ is Cauchy means that $(x_n)$ also converges to some point $x$, since for every $\epsilon >0$ we can find $n_0$ such that for $n \geq n_0$ we have that $|x_n - x| < \epsilon$. I think that in my next step I should show that $x\in X$, but I don't really know how.

Question: How do I show that $(X,d)$ is complete? Or, if I'm on the right track: how do I show that $x \in (X,d)$?

Answer 1

If it is Cauchy, then there exists $N$ such that if $n>N$, we have that $d(x_n,x_m) <1/2$ for all $n,m>N$. Since the metric is discrete, this imposes that the sequence is constant after $N$.

Answer 2

For all $\epsilon>0$ there exists $n_0$ such that for all $n,m\geq n_0$ we have $d(x_n,x_m)<\epsilon$. In particular for $\epsilon=1/2$ there exists $N$ such that for all $n,m\geq N$ $$d(x_n,x_m)<1/2$$ But since the distance can take values only $0$ and $1$, this means that for all $n,m\geq N$ $$d(x_n,x_m)=0.$$ Thus $(x_n)\rightarrow x_N$.