# Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is Cauchy sequence

Let $$(a_n)$$ and $$(b_n)$$ be Cauchy sequences of rationals. Then $$(a_nb_n)$$ is a Cauchy sequence.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

Lemma: If $$(a_n)$$ is a Cauchy sequence of rationals, then for all $$n \in \Bbb N$$, $$|a_n| < A$$ for some $$A \in \Bbb Q$$.

By lemma, there exists $$A$$ such that $$|a_n| < A$$ and $$|b_n| < A$$ for all $$n \in \Bbb N$$.

For a given $$\epsilon >0$$, take an integer $$N$$ such that $$|b_n-b_m|<\dfrac{\epsilon}{2A}$$ and $$|a_n-a_m|<\dfrac{\epsilon}{2A}$$ for all $$n>N$$.

\begin{align} |a_nb_n-a_mb_m| &=|a_n(b_n-b_m) + b_m(a_n-a_m)|\\ &\le |a_n(b_n-b_m)| + |b_m(a_n-a_m)|\\ &= |a_n||b_n-b_m| + |b_m||a_n-a_m|\\ &< A\dfrac{\epsilon}{2A}+ A\dfrac{\epsilon}{2A}\\ &=\epsilon \end{align}

Hence $$(a_nb_n)$$ is a Cauchy sequence.

• It looks fine to me. – José Carlos Santos Jan 30 '19 at 16:00
• By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational. – Mars Plastic Jan 30 '19 at 16:06

Just because there is always room for improvement, I have a tiny nit-pick. When you take $$N$$, you could mention explicitly that it is guaranteed to exist because $$a_n$$ and $$b_n$$ are both Cauchy.