FEEL FREE TO SKIP THIS FIRST PARAGRAPH:(First of all, this is not for marks. This is a statement that the professor has made without proof, and was not included in the homework for this unit. A little background on myself: I am quite rusty in real analysis. It has been a few years since I took Elementary Real Analysis, and my proof writing is quite rusty. I hope that some experienced mathematical thinkers here will point me in the right direction. Thank you in advance.)
QUESTION STARTS HERE: I am trying to prove the following: "First, we define
$$Q=\{(x_{n})_{n \geq 1}:Cauchy \ sequence \ in \ \mathbb{Q}\}$$
That is, Q is the set of cauchy sequences in $ \mathbb{Q}$. We want to prove that if $(x_{n})_{n \geq 1}, \ (y_{n})_{n \geq 1} \in Q$, then
$$(x_{n})_{n \geq 1} \mathcal{R} (y_{n})_{n \geq 1} (\mathcal{R} \ \ is \ \ an \ \ equivalence \ \ relation) \Longleftrightarrow The \ \ two \ \ sequences \ \ are \ \ equivalent$$
We define equivalence in the following way
$$ (*) \ \ : \forall \ \varepsilon>0, \ \exists \ N \in \mathbb{N} \ such \ that \ d(x_{n},y_{n}) \leq \varepsilon, \ n \in \mathbb{N} $$
Proof: Let us assume that (*) holds. Then we have $|x_{n}-y_{n}| \leq \varepsilon$ for $\varepsilon >0$ and n greater than some large enough $N \in \mathbb{N}$. We will now show that the following Cauchy Sequence in the rationals $(x_{n})_{n \geq 1}, \ (y_{n})_{n \geq 1}$ form an equivalence relation.
(I.) We show reflexivity. Let $ (x_{n})_{n\geq1} = (x_{n}^{*})_{n\geq1} $ . Then $ d(x_{n},x_{n}^{*}) = |x_{n}-x_{n}^{*}|=0 \leq \varepsilon $ and $d(x_{n}^{*},x_{n}) = |x_{n}^{*}-x_{n}|=0 \leq \varepsilon $. Therefore, reflexivity holds.
(II.) We now show symmetry. We have $d(x_{n},y_{n})=|x_{n}-y_{n}|<\varepsilon$ and $d(y_{n},x_{n})=|y_{n}-x_{n}|<\varepsilon$. Therefore, symmetry holds.
(III.) We now show transitivity. Let us assume we have a third cauchy sequence in $\mathbb{Q}$, $ (z_{n})_{n \geq 1} $. Let us also assume that $|x_{n}-y_{n}| \leq \frac{\varepsilon}{2}$ and $|y_{n}-z_{n}| \leq \frac{\varepsilon}{2}$. Then we can write the following: $$ | (x_{n}-y_{n}) + (y_{n}-z_{n}) | \leq | (x_{n}-y_{n})| + |(y_{n}-z_{n}) | \leq \frac{2\varepsilon}{2} = \varepsilon $$ We also know that $$ | (x_{n}-y_{n}) + (y_{n}-z_{n}) | =|x_{n}-z_{n}| $$ So we have $ |x_{n}-z_{n}| \leq \varepsilon $. So transitivity holds.
So as far as I'm concerned, the $(\Longleftarrow)$ part is done. Now I must prove that $(x_{n})_{n \geq 1} \mathcal{R} (y_{n})_{n \geq 1} \Longrightarrow The \ \ two \ \ sequences \ \ are \ \ equivalent$. However, is this result trivial? If not, can I get a point in the right direction to start this part of the proof. Also, how was the first part of the proof? Thanks in advance!
I sense some confusion here. The relation you are considering has been shown by you to meet the three requirements for an equivalence relation (purportedly). If this is so, then you are done. There is no reverse direction. It's not that it's trivial. It doesn't exist. What is the relation $\mathcal R$ other than the equivalence you are considering?