Proof verification for $x_ny_n$ converges if and only if $x_n,y_n$ both converge.

Question

The first part of the statement is false, because if we let $x_n=(-1)^n$ and $y_n=(-1)^{n+2}$, which are both divergent sequences, but $x_ny_n=(-1)^n(-1)^{n+2}=(-1)^{2n+2}$ is convergent to 1.

But I think the second part of this statement is true. To prove the second part, suppose $x_n \rightarrow L$ and $y_n \rightarrow M$ for some $L,M \in \mathbb{R}$. Now, let $N \in \mathbb{N}$ such that $n \geq N$ implies $|x_n-L| \lt \epsilon$ and $|y_n-M| \ \lt \epsilon$ where $N$ corresponds to $\epsilon \gt 0$.

$|(x_n-L)(y_n-M)|=|(x_n-L)||(y_n-M)| \lt \epsilon^2 \lt \epsilon$, which is true when $0 \lt \epsilon \lt 1$.

Expanding the two factors,

$|(x_n-L)(y_n-M)| = |x_ny_n-x_nM-y_nL+LM|=$

$|x_ny_n-x_nM-y_nL+LM-2LM+2LM|$ =

$|x_ny_n-LM-x_nM+LM-y_nL+LM|=$

$|x_ny_n-LM-M(x_n-L)-L(y_n-M)| \geq |x_ny_n-LM| - |M(-x_n+L)|-|L(-y_n+M)| =\\ |x_ny_n-LM| - |M||(-x_n+L)|-|L||(-y_n+M)| $

Since $|x_n-L| \lt \epsilon$ and $|y_n-M| \lt \epsilon$, the last two terms will go to $0$. Thus, we are left with

$|x_ny_n-LM| \leq |(x_n-L)(y_n-M)| \lt \epsilon^2 \lt \epsilon$.

Only thing that bothers me is my restriction for values of $\epsilon.$ To be exact, $\epsilon \geq 1$ should also be allowed, but for that case, my proof won't work. But I thought since $\epsilon$ is some number that is very very small, I assumed I could do such restriction. Is this okay to do?

Edit: Also, I used $|a-b| \geq |a|-|b|$, is this true?

Answer 1

Your proof is correct. However, I would replace

Since $|x_n-L| \lt \epsilon$ and $|y_n-M| \lt \epsilon$, the last two terms will go to $0$. Thus, we are left with $|x_ny_n-LM| \leq |(x_n-L)(y_n-M)| \lt \epsilon^2 \lt \epsilon$.

by

This shows that $$\begin{align} |x_ny_n-LM|&\leq |(x_n-L)(y_n-M)|+ |M||(-x_n+L)|+|L||(-y_n+M)|\\ &<\epsilon+|M|\epsilon+|L|\epsilon=(1+|M|+|N|)\epsilon \end{align}$$ Thus, if if we repeat the calculation with $\epsilon$ replaced by $\frac{\epsilon}{1+|M|+|N|}$, we are left with $|x_ny_n-LM| \lt \epsilon$.

Alternatively, instead of taking $N$ such that $n>N$ implies $$|x_n-L|<\epsilon,\quad |y_n-M|<\epsilon,$$ you could take $N$ such that $n>N$ implies $$|x_n-L|<\frac{\epsilon}{1+|M|+|N|},\quad |y_n-M|<\frac{\epsilon}{1+|M|+|N|}.$$ Then, the end of the proof could be

This shows that $$\begin{align} |x_ny_n-LM|&\leq |(x_n-L)(y_n-M)|+ |M||(-x_n+L)|+|L||(-y_n+M)|\\ &<\frac{\epsilon}{1+|M|+|N|}+|M|\frac{\epsilon}{1+|M|+|N|}+|L|\frac{\epsilon}{1+|M|+|N|}=\epsilon \end{align}$$

Yes, it is ok to do that restriction because if $|x_ny_n-LM|<\epsilon$ for some $\epsilon<1$ then $|x_ny_n-LM|<\varepsilon$ for all $\varepsilon\geq 1$. In general, in the delta-epsilon proofs, the values of $\epsilon$ can always be restricted to some small interval of the form $(0,k)$ (in your case, $k=1$).

With respect to the reverse triangle inequality, see this.