Convergence of distance between Cauchy sequences If $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences in $(X,d)$, how do I show that $d(a_n , b_n)$ converges?
 A: Since $$d(a_m,b_m)\leq d(a_m,a_n)+d(a_n,b_n)+d(b_n,b_m)$$ we have $$d(a_m,b_m)-d(a_n,b_n)\leq d(a_m,a_n)+d(b_n,b_m)$$
Since $$d(a_n,b_n)\leq d(a_n,a_m)+d(a_m,b_m)+d(b_m,b_n)$$ we have $$d(a_n,b_n)-d(a_m,b_m)\leq d(a_n,a_m)+d(b_m,b_n)$$
Hence $$0\leq |d(a_m,b_m)-d(a_n,b_n)|\leq d(a_m,a_n)+d(b_m,b_n)\to 0$$ as $m,n\to +\infty$.
This means that $(d(a_n,b_n))$ is Cauchy in $\mathbb R$ so that it converges by the completeness of $\mathbb R$.
A: Hint: For each $n$, let $$c_n:=d(a_n,b_n).$$
This gives a sequence of reals. To show that it is convergent, it suffices to show that it this Cauchy.
A: $|d(a_{n+p}, b_{n+p}) - d(a_n, b_n)| \leqslant |d(a_{n+p}, b_{n+p}) - d(a_{n+p}, b_{n})| + |d(a_{n+p}, b_{n}) - d(a_{n}, b_{n})|$. By the triangular inequality, this is $\leqslant d(b_n, b_{n+p}) + d(a_n, a_{n+p})$. But $a_n$ and $b_n$ are Cauchy, so $d(a_n, b_n)$ is a real Cauchy sequence, so it converges.
A: The key is that for all $m, n \in \mathbb{N}$,
$$d(a_m, b_m) \leq d(a_m, a_n) + d(a_n, b_n) + d(b_m, b_n)$$
from using triangle inequality twice.
It suffices to show that $\{d(a_n, b_n) \}$ is a Cauchy sequence in $\mathbb{R}$.
Fix $\varepsilon > 0$. $\{ a_n \}$ and $\{b_n\}$ are Cauchy in $(X, d)$, so $\exists N \in \mathbb{N}$ such that $m, n > N \implies d(a_m, a_n) < \varepsilon/2$ and $d(b_m, b_n) < \varepsilon/2$. So for $m, n > N$, if $d(a_m, b_m) \geq d(a_n, b_n)$ then
$$|d(a_m, b_m) - d(a_n, b_n)| = d(a_m, b_m) - d(a_n, b_n) \leq d(a_m, a_n) + d(b_m, b_n) < \varepsilon \text, $$
and the case $d(a_n, b_n) > d(a_m, b_m)$ is the same.
