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Let $(p_n)$ and $(q_n)$ be sequences in the metric space $(X, d)$ and assume that $p_n \rightarrow p \in X$ and $q_n \rightarrow q \in X$. Prove that $d(p_n, q_n)$ converges to $d(p, q)$.

Ok, so using the triangle inequality (and assuming the sequences are Cauchy - but can I do that?), I can prove that the distance does converge, but how do I say it converges to $d(p,q)$ exactly?

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Why don't you look at the sequence d(p, $q_n$) -- what does that converge to? – Betty Mock Dec 4 '13 at 0:48
Assume it converges to $d'\neq d(p,q)$, where $|d'-d(p,q)|=\varepsilon > 0$. Go far enough in the sequences that $d(p_n,p)$, $d(q_n,q)$, and $|d'-d(p_n,q_n)|$ are all less than $\varepsilon/3$. Show a contradiction. – mjqxxxx Dec 4 '13 at 0:49
up vote 2 down vote accepted

The key is to use the reverse triangle inequality. Since $d$ is a distance, you have $d(x,y)\leq d(x,z)+d(z,y)$ for any $x,y,z\in X$. This you can write as $$ d(x,y)-d(z,y)\leq d(x,z). $$ As the roles of $x$ and $z$ can be reversed, you get the reverse triangle inequality $$ |d(x,y)-d(z,y)|\leq d(x,z). $$

Then $$ |d(p_n,q)-d(p,q)|\leq d(p_n,p),\ \ \ |d(p_n,q_n)-d(p_n,q)|\leq d(q_n,q). $$ Using the above and the regular triangle inequality, $$ |d(p_n,q_n)-d(p,q)|\leq |d(p_n,q_n)-d(p_n,q)|+|d(p_n,q)-d(p,q)|\leq d(q_n,q)+d(p_n,p). $$ So $\lim_{n\to\infty}d(p_n,q_n)=d(p,q)$.

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