If $\{x_n\}$ and $\{y_n\}$ are Cauchy then $\left\{\frac{2x_n}{y_n}\right\}$ is Cauchy

Question

Let $S\subseteq \mathbb{R}$, and $\{x_n\}$ and $\{y_n\}$ be two Cauchy sequences in $S$, with $x_n,y_n\geq 10$. Prove that $z_n:=\left\{\frac{2x_n}{y_n}\right\}$ is also Cauchy in $S$. [It is assumed that $z_n\in S$.

My proof:

Since $x_n, y_n$ are Cauchy in $S$, they are convergent in $S$ and thus bounded in $S$. Let $x_n$ converge to real number $L$ and $y_n$ converge to real number $M$. Since $x_n, y_n$ are Cauchy, $\left|x_n-x_m\right|<\varepsilon$ for $N_1(\varepsilon)<n,m$ and $\left|y_n-_m\right|<\varepsilon$ for $N_2(\varepsilon)<n,m$. WLOG, suppose that $y_n > y_m$. Let $N:=\max\{N_1, N_2\}$, then $$\begin{align}\left| \frac{2x_n}{y_n}-\frac{2x_m}{y_m}\right| & =\left|\frac{2x_ny_m-2x_my_n}{y_ny_m}\right| \\&\leq \left|\frac{x_ny_m-x_my_n}{5y_m}\right|\\&< \left| x_n-x_m\frac{y_n}{y_m} \right|\\&\leq \underbrace{\left| x_n-x_m \right|}_{\text{since } \left| y_n-y_m\right| <\varepsilon\implies \frac{y_n}{y_m}\ge 1}\\ &<\varepsilon.\end{align}$$ Hence $z_n$ is Cauchy.

End of proof.

But... I'm somewhat concerned that the argument under the underbrace is too handwavy. Would appreciate some help.

Answer 1

Since they are Cauchy in $S$, they are also Cauchy in $\mathbb R$. Hence they are convergent in $\mathbb R$. Since $y_n\ge10$ we have that $\lim_{n\to\infty}\limits y_n = M\ge 10$, in particular $\lim_{n\to\infty}\limits y_n \not=0$. Hence $z_n=\frac{2x_n}{y_n}$ is convergent in $\mathbb R$, hence it is Cauchy in $\mathbb R$, hence Cauchy in $S$.