# Discrete metric

## Main Question or Discussion Point

how is discrete metric space given by d((x1,x2,....xn)(y1,y2,....yn))=0 if xi=yi else 1
disc
is complete

A metric space is "complete" if and only if every Cauchy sequence converges. And, of course, a Cauchy sequence is one where $\lim_{m,n\rightarrow \infty} d(a_n,a_m)= 0$. Since d(x,y)= 1 for $x\ne y$, in order for that to happen the sequence must be "eventually constant", i.e. for some N, if n,m> N, $a_n= a_m$ and it is easy to show that such a sequence converges.