showing a (B,p) is complete let $(A,d)$ be a metric space which is complete and let $B$ be closed in A. Then prove $(B,p)$ is complete, with $p = d|_{B\times B}$ (i.e. restriction of d onto $B\times B$)
thoughts:
since $B$ is closed in $A$ and $(A,d)$ is complete i figured that $(B,y)$ is complete - from here I know I have to show for $x_n,x_m \in B$,$ p(x_n,x_m) < \epsilon $ for $x,m\geq N$ but I am not sur ehow to deal with the restriction of $d$ on $B\times B$ - I mean, how would I go about evaluating $p(x_n,x_m)$?
edit: progress:
$d|_{YXY} ((x_n,x_m),(y_n,y_m)) = p((x_n,x_m),(y_n,y_m)) $, with $(x_n,x_m), (y_n,y_m) \in Y$ now if I let $x_n,y_n$ be cauchy I can see that $p<\epsilon$ but no idea how ot prove it

Let $X$ be a complete metric space and $F$ a closed set in $X$. If $x_n$ is a Cauchy sequence in $F$, then it has a limit $x\in X$. Since $F$ is closed, $x\in F$. Hence $F$ is complete.
