2

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?

JOlv
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    @RobertIsrael Any uniformly continuous function of Cauchy sequences is Cauchy. On the metric space $(0, \infty)$, the map $x \mapsto 1/x$ is continuous, but fails to preserve Cauchy sequences, e.g. $(\frac{1}{n})_n$ is Cauchy, but not its image under the above continuous map. – Theo Bendit Aug 09 '21 at 23:25

1 Answers1

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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.

Theo Bendit
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  • Oh, I see! That does make sense, since norms are non-negative. Thank you! – JOlv Aug 09 '21 at 23:40
  • @JOlv You're welcome! Actually, I meant to bring this up: you seem to be confusing normed linear spaces and metric spaces. The function is a metric, not a norm. In the body of your question, you refer to a $0$ point in the space, but metric spaces have no such privileged point called $0$. While normed linear spaces can give you some intuition about metric spaces, it's worth keeping the concepts separate in your head. – Theo Bendit Aug 09 '21 at 23:56