# Prove that every set for the trivial metric is complete.

Given is the trivial metric : $$d(x,y) = \left\{ \begin{array}{ll} 0 & \text{if} & x=y \\ 1 & \text{if} & x \neq y \end{array} \right.$$ I have to proof that for any set $X$, $(X,d)$ is complete. This means that every Cauchyserie has to converge. Say that $(x_n)$ is a Cauchy and that $a$ is it's limit for large $n$. For every $\epsilon > 0$ then there exist a $n_0$ such that for all $n$ greater than $n_0$ : $d(x_n, a) < \epsilon$.

When $x_n \rightarrow a$ it is obvious that $d(x_n, a) < \epsilon$. But I can't seem to prove it for $x \neq y$ because the distance will always be $1$, not depending on how large you take $n_0$.

I am probably overlooking something simple, so if you could push me in the right direction that would be appreciated. Thanks.

You are making a mistake: when you say “and that $a$ is its limit”, you are assuming that there is a limit. You must prove it.
Note that there is natural $n_0$ such that $m,n\geqslant n_0\implies d(x_m,x_n)<1$. But then $d(x_m,x_n)=0$, that is $x_m=x_n$. So, the limit is $x_{n_0}$.
• But how do you know that there is a $n_0$ so that $d(x_m, x_n) < 1$? Same for Tsemo's answer, he choses $1/2$ instead. Commented May 30, 2018 at 13:06
• That the elements of the sequence get closer together further in the sequence. The number $1$ or $1/2$ just seems arbitrary to me, if you take $2$ for example, it wouldn't fit. Commented May 30, 2018 at 13:17
• @WarreG You erroneously wrote "For a random $\epsilon>0$ ..." Should be: "For every $\epsilon>0$ ...", in particular for $\epsilon=1/2$. Commented May 30, 2018 at 13:22
• @WarreG Because with $1$ or $\frac12$ I can extract an useful conclusion. With $2$, I cannot. Commented May 30, 2018 at 13:51
Let $(x_n)$ be a Cauchy sequence, there exists $N$ such that $n,m\geq N$ implies that $d(x_n,x_m)<1/2$, this implies that $x_n=x_m=x_N$ and $(x_n)$ converges.