Problem: Suppose that $\left( x_n \right) $ and $\left( y_n \right) $ are sequences in a metric space $X$ with metric $d$. Let $\left( x_n \right) $ be Cauchy and $\displaystyle \lim_{n \to \infty} d\left( x_n, y_n \right) = 0$. Prove that $\left( y_n \right) $ is Cauchy.
Solution: Let $\varepsilon \gt 0$ be given. We need to show that there exists $n_0 \in \mathbb{N} $ such that for $m, n \geq n_0$ we have $d\left( y_n, y_m \right) \lt \varepsilon $. Since $\left( x_n \right) $ is Cauchy, for the given $\varepsilon $ we will get $n_1 \in \mathbb{N} $ such that for $m,n \geq n_1$
\begin{equation}\label{eq:31Jul2023-1}
d \left( x_n, x_m \right) \lt \frac{\varepsilon }{3}.
\end{equation}
Now we also have $\displaystyle \lim_{n \to \infty} d \left( x_n, y_n \right) = 0$, so there exists $n_2 \in \mathbb{N} $ such that for $n \geq n_2$
\begin{equation}\label{eq:31Jul2023-2}
d \left( x_n, y_n \right) \lt \frac{\varepsilon}{3}.
\end{equation}
Take $n_0 = \max \left\{ n_1, n_2 \right\} $. Now observe that for $n\geq n_0$
\begin{align*}
d \left( y_n, y_m \right) & \leq d \left( y_n, x_n \right) + d \left( x_n, x_m \right) + d \left( x_m, y_m \right) \\
& \lt \frac{\varepsilon }{3} + \frac{\varepsilon }{3} + \frac{\varepsilon }{3} = \varepsilon .
\end{align*}
This proves that $\left( y_n \right) $ is Cauchy.