Let $\epsilon > 0$. Show that if $(a_{n})^{\infty}_{n=1}$ and $(b_{n})^{\infty}_{n=1}$ are eventually $\epsilon$-close, then $(a_{n})^{\infty}_{n=1}$ is bounded iff $(b_{n})^{\infty}_{n=1}$ is bounded.
MY ATTEMPT
If $a_{n}$ and $b_{n}$ are eventually $\epsilon$-close, then there exists a natural number $N\geq 0$ such that $|a_{n} - b_{n}| \leq \epsilon$ for every $n\geq N$. If $a_{n}$ is bounded, one has that $|a_{n}| \leq M$. Consequently, we have that \begin{align*} |b_{n}| \leq |a_{n} - b_{n}| + |a_{n}| \leq \epsilon + M \end{align*} for $n\geq N$. If we take $B = \max\{|b_{1}|,|b_{2}|,\ldots,|b_{N-1}|,\epsilon + M\}$, then $|b_{n}|\leq B$ for every natural $n$, and we are done.
Similar reasoning applies to the converse case.
Could someone verify my arguments?