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.

  • 5
    $\begingroup$ It looks fine to me. $\endgroup$ – José Carlos Santos Jan 30 '19 at 16:00
  • $\begingroup$ By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational. $\endgroup$ – Mars Plastic Jan 30 '19 at 16:06

This looks good. Well done.

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.


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