Let the Cauchy sequences $a_{n}$ and $a'_{n}$ be equivalent. If $b_{n}$ is Cauchy, then prove $a_{n}b_{n}$ is equivalent to $a'_{n}b_{n}$.

Question

Suppose the Cauchy sequences $a_{n}$ and $a'_{n}$ are equivalent. If $b_{n}$ is a Cauchy sequence, then prove that $a_{n}b_{n}$ is equivalent to $a'_{n}b_{n}$.

MY ATTEMPT

We have to prove that, for every positive rational $\varepsilon > 0$, there is a natural number $N$ such that \begin{align*} |a_{n}b_{n} - a'_{n}b_{n}| \leq\varepsilon \end{align*} whenever $n\geq N$. Indeed, let us start by noticing that \begin{align*} |a_{n}b_{n} - a'_{n}b_{n}| = |b_{n}||a_{n} - a'_{n}| \leq B|a_{n}-a'_{n}| \end{align*}

This is because $b_{n}$ is a Cauchy sequence, therefore it is bounded by some positive rational $B > 0$.

Moreover, since $a_{n}$ and $a'_{n}$ are equivalent, for every $\varepsilon/B > 0$, there is a natural number $N \geq 1$ such that \begin{align*} |a_{n} - a'_{n}| \leq \varepsilon/B \end{align*} whenever $n\geq N$. Thus we conclude that, for every $\varepsilon > 0$, there is a natural number $N\geq 1$ such that \begin{align*} |a_{n}b_{n} - a'_{n}b_{n}| \leq B|a_{n} - a'_{n}| \leq B\times\frac{\varepsilon}{B} = \varepsilon \end{align*} whenever $n\geq N$, and we are done.

Could someone double-check my arguments?

EDIT

Two sequences $a_{n}$ and $b_{n}$ are equivalent iff for every positive rational $\varepsilon > 0$, there is a natural number $N\geq 1$ such that \begin{align*} |a_{n} - b_{n}|\leq\varepsilon \end{align*} whenever $n\geq N$.

Answer 1

The essential part is using the fact $(b_n)$ must be bounded. Yes, your arguments are correct.