Understanding Analysis · Chapter 2 · §2.6

The Cauchy Criterion

Exercise 2.6.4

Let $(a_{n})$ and $(b_{n})$ be Cauchy sequences. Determine whether $c_{n} = |a_{n} - b_{n}|$ is a Cauchy sequence.

Proof.
To show $c_{n} = |a_{n} - b_{n}|$ is a Cauchy sequence, we need to show that for every $\epsilon > 0$ , there exists $N \in \mathbb{N}$ such that $m, n \geq N$ implies

||a_{n} - b_{n}| - |a_{m} - b_{m}|| < \epsilon

We use the reverse triangle inequality

\begin{aligned} ||a_{n} - b_{n}| - |a_{m} - b_{m}|| &\leq |a_{n} - b_{n} - (a_{m} - b_{m})| \\ &\leq |a_{n} - a_{m} - b_{n} + b_{m}| \\ &\leq |(a_{n} - a_{m}) + (b_{m} - b_{n})| \end{aligned} \tag{1}

Using the triangle inequality

|(a_{n} - a_{m}) + (b_{m} - b_{n})| \leq |a_{n} - a_{m}| + |b_{m} - b_{n}| \tag{2}

We are given that $(a_{n})$ and $(b_{n})$ are Cauchy sequences. By the definition of Cauchy, there exists an $N_{a}$ and $N_{b}$ where the terms in $(a_{n})$ and $(b_{n})$ get arbitrarily close. We then choose $N = \text{max}\{N_{a}, N_{b}\}$ such that

|a_{n} - a_{m}| < \frac{\epsilon}{2} \quad \text{and} \quad |b_{m} - b_{n}| < \frac{\epsilon}{2}

for every $m, n \geq N$ .

Then we can combine with the LHS of (1) and substitute into the RHS of (2)

||a_{n} - b_{n}| - |a_{m} - b_{m}|| \leq |(a_{n} - a_{m}) + (b_{m} - b_{n})| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

We have found an $N$ where $||a_{n} - b_{n}| - |a_{m} - b_{m}|| < \epsilon$ for every $m, n \geq N$ . That means $c_{n} = |a_{n} - b_{n}|$ is Cauchy. $\blacksquare$

Stephen Abbott, Understanding Analysis, 2nd edition.