Hi everyone I'm having a bad time with two questions in the Analysis book of Terry Tao. I finally finished one of the exercises and I'm wondering if the next reasoning is correct or maybe needs some changes:
Definitions:
Two sequence are equivalence $\iff$ $(\forall \varepsilon \in \mathbb{Q}^+\,) ( \, \exists N\in \mathbb{N}\,) \text{ s.t. }(\, \forall n \ge N\, (|a_n-b_n|\le \varepsilon)\,) $
(where $\mathbb{Q}^+$ is a positive rational number).
Exercise: Show that if $\langle a_n \rangle_{n=1}^{\infty}$ and $\langle b_n \rangle_{n=1}^{\infty}$ are equivalences secuences of rationals. Then $\langle a_n \rangle_{n=1}^{\infty}$ is a Cauchy sequence if and only if $\langle b_n \rangle_{n=1}^{\infty}$ is a Cauchy sequence.
Proof:
We suppose that $\langle a_n \rangle_{n=1}^{\infty}$ is a Cauchy sequence, and also we may assume that $\langle a_n \rangle_{n=1}^{\infty}$ and $\langle b_n \rangle_{n=1}^{\infty}$ are equivalent sequences; we wish to show that $\langle b_n \rangle_{n=1}^{\infty}$ is a Cauchy sequence.
Let $\, \varepsilon $ be an arbitrary positive integer, we shall show that there is some $N \in \mathbb{N}$ such that $|b_j-b_k| \le \varepsilon$ for all $\,j,k \ge N$. Let $\gamma$ be a positive rational number such that $\gamma < \varepsilon$. Since $\langle a_n \rangle _{n = 1}^{\infty}$ is a Cauchy sequence it follows that there is some $N'\in \mathbb{N}$ such that $|a_{j'}-a_{k'} |\le \varepsilon-\gamma$ for all $j',k' \ge N'$.
Now we set $|a_n-b_n|\le \frac{\gamma}{2}\,$ for all $n \ge M$. So either $M>N'$ or $\,M\le N'$ by the ordering of natural numbers.
If $M>N'$ we choose $j',k'\ge M$ and then, we have that:
$|b_{j'} -b_{k'}|-|a_{j'} -a_{k'}|\le|(a_{j'} -a_{k'})- (b_{j'} -b_{k'})| \le |a_{j'} -b_{j'}|+|a_{k'} -b_{k'}|\le \gamma$
And so we have that $|b_{j'} -b_{k'}|\le \gamma +|a_{j'} -a_{k'}| \le \varepsilon$, it follows that $|b_{j'} -b_{k'}|\le \varepsilon$ for all $j',k'\ge M$ as desired.
On the other hand if $\,M \le N'$ we choose a $n\ge N'$ and a similar argument give us that $|b_{j'} -b_{k'}|\le \varepsilon$ for all $j',k'\ge N'$ as desired.
So in either case, $|b_{j} -b_{k}|\le \varepsilon\,$ for all $\,j,k \ge N$ and hence the sequence $\langle b_n \rangle_{n=1}^{\infty}$ is a Cauchy sequence.
To conclude note that the converse may be disposed with no additional work, by applying the same argument with the roles of $\langle a_n \rangle_{n=1}^{\infty}$ and $\langle b_n \rangle_{n=1}^{\infty}$ interchanged, i.e., where $\langle b_n \rangle_{n=1}^{\infty}$ is a Cauchy sequence.
Thanks in advance as usual.
Best Answer
Let $\epsilon>0$ be give. With loss of generality, we may assume $\epsilon$ is rational. Suppose $a_n$ is a Cauchy sequence and $b_n$, $a_n$ are equivalent.
Choose $N$ such that $|b_k-a_k| < \frac{1}{3} \epsilon$ and $|a_n-a_m| < \frac{1}{3} \epsilon$ whenever $k,m,n \ge N$. We can do this because $a_n$ is Cauchy, and $a_n,b_n$ are equivalent.
Then if $n,m \ge N$, we have $|b_n-b_m| \le |b_n-a_n|+|a_n-a_m|+|a_m-b_m| \le \frac{1}{3} \epsilon + \frac{1}{3} \epsilon + \frac{1}{3} \epsilon = \epsilon$. Hence $b_n$ is Cauchy.
Reversing the roles of $a_n,b_n$ finishes the proof.