# show sequence converges

If $(x_n)$ and $(y_n)$ are Cauchy sequences in a metric space X with metric d, how would I show the sequence $(d(x_n,y_n))$ converges?

I'm supposed to use $d(x_n,y_n)\leq d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)$ If I subtract by $d(x_m,y_m)$ on both sides:

$d(x_n,y_n)-d(x_m,y_m)\leq d(x_n,x_m)+d(y_m,y_n)$

Since $d(x_n,x_m)$ and $d(y_m,y_n)$ are Cauchy, then each can be less than $\frac{\epsilon}{2}$ which makes the left side $\leq \epsilon$. How would I show it converges though?

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HINT: Show that $\langle d(x_n,y_n):n\in\Bbb N\rangle$ is a Cauchy sequence in $\Bbb R$, and use the fact that $\Bbb R$ is complete: every Cauchy sequence in $\Bbb R$ converges.
 Thanks, how would show it's a Cauchy sequence in R particularly, though? – Alti Oct 18 '12 at 19:47 @Alti: By showing that it satisfies the definition of a Cauchy sequence: for each $\epsilon>0$ there is an $n_0\in\Bbb N$ such that $|d(x_m,y_m)-d(x_n,y_n)|<\epsilon$ whenever $m,n\ge n_0$. You’ve already done much of the work. – Brian M. Scott Oct 18 '12 at 19:52 Great, thank you. – Alti Oct 18 '12 at 20:01
The triangle inequality gives $$-d(y_n,y_m)\le d(x_n,y_n)-d(x_n,y_m)\le d(y_n,y_m)$$ and $$-d(x_n,x_m)\le d(x_m,y_m)-d(x_n,y_m)\le d(x_n,x_m)$$ Subtracting yields $$|d(x_n,y_n)-d(x_m,y_m)|\le d(x_n,x_m)+d(y_n,y_m)$$