We are tasked with showing that if two sequences $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ are equivalent sequences, then $(a_n)_{n \in \mathbb{N}}$ is Cauchy if and only if $(b_n)_{n \in \mathbb{N}}$ is Cauchy.

Definitions:

1. Equivalent sequences: Two sequences $(a_n)$ and $(b_n)$ are said to be equivalent if: $\lim_{n \to \infty} |a_n - b_n| = 0.$ This means that as $n \to \infty$, the terms of the two sequences become arbitrarily close to each other.

2. Cauchy sequence: A sequence $(a_n)$ is called a Cauchy sequence if for every $\epsilon > 0$, there exists an integer $N$ such that for all $m, n > N$, $|a_n - a_m| < \epsilon.$ This means that the terms of the sequence become arbitrarily close to each other as $n$ and $m$ grow large.

Proof Outline:

We aim to prove that:

• If $(a_n)$ is Cauchy, then $(b_n)$ is also Cauchy, and vice versa.

Direction 1: If $(a_n)$ is Cauchy, then $(b_n)$ is Cauchy.

Assume $(a_n)$ is a Cauchy sequence. This means that for every $\epsilon > 0$, there exists $N_1 \in \mathbb{N}$ such that for all $m, n > N_1$, $|a_n - a_m| < \frac{\epsilon}{3}.$ Since $(a_n)$ and $(b_n)$ are equivalent, we have: $\lim_{n \to \infty} |a_n - b_n| = 0.$ So, for this $\epsilon > 0$, there exists $N_2 \in \mathbb{N}$ such that for all $n > N_2$, $|a_n - b_n| < \frac{\epsilon}{3}.$ Similarly, for all $m > N_2$, $|a_m - b_m| < \frac{\epsilon}{3}.$

Now, for all $m, n > \max(N_1, N_2)$, we can estimate $|b_n - b_m|$ using the triangle inequality: $|b_n - b_m| \leq |b_n - a_n| + |a_n - a_m| + |a_m - b_m|.$ Using the bounds we established earlier: $|b_n - b_m| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon.$ Thus, $(b_n)$ is also a Cauchy sequence.

Direction 2: If $(b_n)$ is Cauchy, then $(a_n)$ is Cauchy.

The proof for this direction is analogous to the first direction. Assume $(b_n)$ is a Cauchy sequence. Then for every $\epsilon > 0$, there exists $N_3 \in \mathbb{N}$ such that for all $m, n > N_3$, $|b_n - b_m| < \frac{\epsilon}{3}.$ Since $(a_n)$ and $(b_n)$ are equivalent, there exists $N_4 \in \mathbb{N}$ such that for all $n > N_4$, $|a_n - b_n| < \frac{\epsilon}{3}.$ Similarly, for all $m > N_4$, $|a_m - b_m| < \frac{\epsilon}{3}.$

Now, for all $m, n > \max(N_3, N_4)$, using the triangle inequality again: $|a_n - a_m| \leq |a_n - b_n| + |b_n - b_m| + |b_m - a_m|.$ Using the bounds we established: $|a_n - a_m| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon.$ Thus, $(a_n)$ is also a Cauchy sequence.

Conclusion:

Since we have shown both directions, we conclude that if $(a_n)$ and $(b_n)$ are equivalent sequences, then $(a_n)$ is Cauchy if and only if $(b_n)$ is Cauchy.

---

Would you like more details or have any questions?

Here are five related questions for further exploration:

1. How do equivalent sequences behave with respect to limits?
2. Can two equivalent sequences converge to different limits?
3. What is an example of two sequences that are equivalent but not identical?
4. How do Cauchy sequences relate to convergent sequences in metric spaces?
5. Can a sequence be Cauchy but not converge in every space?

Tip: In any metric space, every Cauchy sequence converges if the space is complete.