What I am trying to prove is the following:
Let $(a_i)_{i = 1}^\infty$ and $(b_i)_{i = 1}^\infty$ be Cauchy sequences of rationals. Then the sequence $(a_ib_i)_{i = 1}^\infty$ is a Cauchy sequence of rationals.
My attempt at proving it is as follows.
Let $\varepsilon, \delta>0$ be rationals. Now, there exists an $N\ge 1$ such that for all $i, j\ge N$, we have $\lvert a_i-a_j\rvert, \lvert b_i-b_j\rvert\le\delta$. Also, since Cauchy sequences are bounded, there exists a rational $M\ge \delta$ such that for each $i\ge 1$, we have $\lvert a_i\rvert, \lvert b_i\rvert\le M$.
Let $i, j\ge N$. Then
\begin{align*}
\lvert a_ib_i-a_jb_j\rvert&\le\delta\lvert b_i\rvert+\delta\lvert a_i\rvert+\delta^2\\
&\le\delta(2M+\delta)\\
&\le3M\delta.
\end{align*}
Now, $3M\delta<\varepsilon\impliedby\delta<\varepsilon/(3M)$ since $M>0$. But there does exist a rational $\tilde\delta$ such that $0<\tilde\delta<\varepsilon/(3M)$. Hence we are done.
Is my proof correct? I am in doubt because I am confused by the quantifiers.
Note that in the above I am using the result that if $x, y, z, w$ are rationals such that $\lvert x-y\rvert\le\varepsilon$ and $\lvert w-z\rvert\le\delta$, then we have $\lvert xz-yw\rvert\le\varepsilon\lvert z\rvert+\delta\lvert x\rvert+\varepsilon\delta$, which has already been proven before in the text I'm following.
Question: In order for the above proof to work, doesn't one also need to ensure that $\tilde\delta\le M$?
You can write $a_ib_i-a_jb_j$ as $a_i(b_i-b_j)+b_j(a_i-a_j) $ So $$|a_i b_i - a_jb_j | \leq |a_i ||b_i-b_j|+ |b_j||a_i-a_j|$$ then you can just use the fact that Cauchy sequences are bounded.