More specifically:
Let $\{x_n\}_{n \in \mathbb{N}}$ and $\{y_n\}_{n \in \mathbb{N}}$ be Cauchy sequences in a metric space $\langle M, \rho \rangle$. Prove that $\{\rho(x_n, y_n)\}_{n \in \mathbb{N}}$ is Cauchy in $\langle \mathbb{R}, \lvert \cdot \rvert_1 \rangle$.
My attempt:
$\{x_n\}_{n \in \mathbb{N}}$ and $\{y_n\}_{n \in \mathbb{N}}$ being Cauchy implies $\rho(x_{m_{1}}, y_{n_{1}}) < \varepsilon_{1}$ and $\rho(x_{m_2}, y_{n_2}) < \varepsilon_2$, where $m_1, m_2, n_1, n_2 \in \mathbb{N}$ and $\varepsilon_1, \varepsilon_2 \in \mathbb{R}^{+}$. We need to show $\lvert \rho(x_n, y_n) – \rho(x_m, y_m) \rvert_1 < \varepsilon$ for some $m, n \geq N \in \mathbb{N}$ and $\varepsilon > 0$.
This is where I run into trouble. I tried using the Reverse Triangle Inequality to bound $\rho(x_m, y_m)$ below by $\lvert \rho(x_m, 0) – \rho(y_m, 0) \rvert_1$ and $\rho(x_n, y_n)$ below by $\lvert \rho(x_n, 0) – \rho(y_n, 0) \rvert_1$, similar to this question.
I don’t really know where to go from here, though. The difference $\lvert \rho(x_n, y_n) – \rho(x_m, y_m) \rvert_1$ is not bounded below by the difference of the lower bounds justified by the RTI, and I don’t know how I can use $\{x_n\}_{n \in \mathbb{N}}$ and $\{y_n\}_{n \in \mathbb{N}}$ being Cauchy to say anything about $\lvert \rho(x_n, y_n) – \rho(x_m, y_m) \rvert_1$.
Any help?
Best Answer
You can use the quadrilateral inequality: $$|\rho(a, b) - \rho(c, d)| \le \rho(b, c) + \rho(a, d).$$ This inequality can be proven using triangle inequality. This inequality gives us $$|\rho(x_n, y_n) - \rho(x_m, y_m)| \le \rho(x_n, x_m) + \rho(y_n, y_m),$$ where the latter two can be made arbitrarily small.