Proof check - If two sequences A and B are equivalent, then the first one is a Cauchy sequence if and only if the second one is a Cauchy sequence too

Could someone check this proof?

Definitions:

Equivalence: Two sequences are equivalent iff, for any $\varepsilon > 0$,there exists an $N$ such that for $n \geq N$, $|a_n - b_n| < \varepsilon$.

Cauchy Sequence: $(b_n)_{n=1}^\infty$ is a Cauchy sequence iff, for any $\varepsilon > 0$, there exists an $N$ such that $j,k \geq N$ implies $|a_j - a_k| < \varepsilon$.

Proposition: Show that, if two sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent, then $(a_n)_{n=1}^\infty$ is a Cauchy sequence if and only if $(b_n)_{n=1}^\infty$ is a Cauchy sequence too.

Proof:

Let $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ be equivalent and assume $(a_n)_{n=1}^\infty$ is a Cauchy sequence.

As both sequences are equivalent, for some n and for any $\varepsilon$ , we have

$$|a_n - b_n| + | b_{n+1} - a_{n+1}| + |a_{n+1} - a_n | < \varepsilon$$

By the triangle inequality theorem,

$$|a_n - b_n + b_{n+1} - a_{n+1} + a_{n+1} - a_n| = |b_{n+1} - b_n| <\varepsilon$$

We could show that, if $(b_n)_{n=1}^\infty$ is a Cauchy sequence, then $(a_n)_{n=1}^\infty$ is a Cauchy sequence too, by a reciprocal argument.

$\blacksquare$

Suppose $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are two equivalent sequences, and let $\varepsilon>0$ be given. Since $\frac{1}{3}\varepsilon>0$, there is a positive integer $N_o$ such that \begin{equation} |a_n - b_n| < \frac{1}{3}\varepsilon \,\text{ whenever }\, n\geq N_o. \end{equation}

Without loss of generality, we shall only assume $(a_n)_{n=1}^\infty$ is a Cauchy sequence. Since $(a_n)_{n=1}^\infty$ is Cauchy, there is a positive integer $N_1$ such that \begin{equation} |a_m - a_n| < \frac{1}{3}\varepsilon \,\text{ whenever }\, m,n \geq N_1. \end{equation} We select $N=\max\{N_o, N_1\}$. So if $m,n \geq N$, then \begin{aligned}|b_m-b_n| & = |b_m-a_m+a_m-a_n + a_n - b_n| \\& \leq |b_m-a_m|+|a_m-a_n|+|a_n-b_n| \\& < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\varepsilon. \end{aligned}

I have a few remarks:

1. When you wrote “for some $n$ and for some $\varepsilon$”, you should have written that, for every $\varepsilon$ there is some $n$.
2. In order to prove that$$|a_n - b_n| + | b_{n+1} - a_{n+1}| + |a_{n+1} - a_n | < \varepsilon,$$you need, not only that the sequences are equivalent, but also that $(a_n)_{n\in\mathbb N}$ is a Cauchy sequence.
• Does the OP not assume 2? – Theoretical Economist Jan 15 '18 at 23:32
• @TheoreticalEconomist No. He claimed that it is true and it is easy to prove. I assumed that the OP knew how to prove it. – José Carlos Santos Jan 15 '18 at 23:35
• yes, I think I see what you mean now. I should’ve read more carefully. :) – Theoretical Economist Jan 15 '18 at 23:40
• As written in the proposition, first of, we assume the sequences are equivalent, then we assume $(a_n)_{n=1}^\infty$ is Cauchy and then we prove $(b_n)_{n=1}^\infty$ is also Cauchy. And, in the reciprocal argument, we assume $(b_n)_{n=1}^\infty$ is Cauchy, and then we prove $(a_n)_{n=1}^\infty$ too. Am I right? – DunhoClark Jan 15 '18 at 23:41
• @DunhoClark Yes, you are. – José Carlos Santos Jan 15 '18 at 23:43

Critique

To show that $b_n$ is Cauchy, you must show that it holds for all $j,k \geq N$: you only showed it for $j,j+1\geq N$. You actually also need to use $N$.

It is also good to mention that each term is smaller than $\frac{\varepsilon}{3}$, so then the $\varepsilon$ bound you have written is justified.

We could show that, if $(b_n)_{n=1}^\infty$ is a Cauchy sequence, then $(a_n)_{n=1}^\infty$ is a Cauchy sequence too, by a reciprocal argument.

This is correct.

Hints

A big step forward is to use $N=\max\{N_1,N_2\}$, where $N_1$ comes from the definition of equivalent sequences, and $N_2$ from the definition of Cauchy sequence (both ending with $|\ldots|<\frac{\varepsilon}{3}$). So if $n,m\geq N$, then both these definitions hold (because we have $n,m\geq N_1$ and $n,m\geq N_2$). Now you can finish the proof with your reasoning.