Given that $x_n$ and $y_n$ are Cauchy sequences in $\mathbb{R} $, prove that $x_n y_n$ is Cauchy without the use of the Cauchy theorem stating that Cauchy $\Rightarrow$ convergence.
Attempt: Without that condition on not been able to use the theorem, the question becomes trivial. Instead:
For all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for $n,m \geq N, -\frac{\epsilon}{2} \leq x_n – x_m \leq \frac{\epsilon}{2}$ and similar statment for $y_n$. Multiply the above by $y_n$ and the equivalent statement for $y_n$ by $x_m$. Then add these together. The result is: $$|x_ny_n – x_my_m| < \frac{\epsilon}{2}(x_m + y_n)$$ I have proved in a previous question that $x_n + y_n$ is Cauchy so could I apply that here and say for $n,m \geq N$, $x_n + y_n$ is Cauchy and hence convergent so tends to a finite limit for $n,m \geq N$. This would mean my upper bound is a multiple of $\epsilon$ and since $\epsilon$ is arbritarily small, so is this upper bound. Hence Cauchy.
I don't think this would warrant a full proof in any case since by multiplying by $x_m$ and $y_n$, I am assuming they are positive so as to not reverse the inequality signs. Nonetheless, I would appreciate some feedback on what I have done.
Many thanks
Best Answer
HINT: You can make the basic idea work by arranging matters a bit differently. Start with
$$x_ny_n-x_my_m=(x_ny_n-x_ny_m)+(x_ny_m-x_my_m)\;,$$
supply and manipulate absolute values appropriately, and use the fact that a Cauchy sequence is bounded. (Note that boundedness can be proved easily without actually showing convergence.)