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If $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences in $(X,d)$, how do I show that $d(a_n , b_n)$ converges?

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It isn’t considered polite to tell people to ‘show you’ something. – Haskell Curry Mar 17 '13 at 23:15
up vote 2 down vote accepted

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 $\mathbb R$.

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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.

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$|d(a_{n+p}, b_{n+p}) - d(a_n, b_n)| \leqslant |d(a_{n+p}, b_{n+p}) - d(a_{n+p}, b_{n})| + |d(a_{n+p}, b_{n}) - d(a_{n}, b_{n})|$. 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.

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