Let $(a_n)_{n=0}^\infty$ be bounded and $(b_n)_{n=0}^\infty$ be equivalent to $(a_n)_{n=0}^\infty$. Show $(b_n)_{n=0}^\infty$ is bounded.

Question

Let $(a_{n})_{n=0}^{\infty}$ be a sequence of rational numbers which is bounded. Let $(b_{n})_{n=0}^{\infty}$ be another sequence of rational numbers which is equivalent to $(a_{n})_{n=0}^{\infty}$. Show that $(b_{n})_{n=0}^{\infty}$ is also bounded.

MY ATTEMPT

Since $a_{n}$ and $b_{n}$ are equivalent, for every positive rational $\epsilon > 0$ there is a natural $N \geq 1$ such that \begin{align*} |a_{n} - b_{n}| \leq \epsilon \end{align*}

for $n\geq N$. In particular, when $\epsilon = 1$, one has that

\begin{align*} |a_{n} - b_{n}|\leq 1 \end{align*}

for $n \geq n_{0}$. Moreover, since $a_{n}$ is bounded, this means that $|a_{n}| \leq M$, for some rational $M > 0$.

Consequently, one has for $n\geq n_{0}$ that

\begin{align*} |b_{n}| = |b_{n} - a_{n} + a_{n}| \leq |a_{n} - b_{n}| + |a_{n}| \leq 1 + M \end{align*}

If we take $B = \max\{|b_{1}| + |b_{2}| + \ldots + |b_{n_{0}}|, 1 + M\}$, we conclude that $|b_{n}|\leq B$ for every natural, as desired.

Can someone double-check my solution?

Answer 1

What you did looks correct. I have just one minor point at the end where you state

$$B = \max\{|b_{1}| + |b_{2}| + \ldots + |b_{n_{0}}|, 1 + M\} \tag{1}\label{eq1A}$$

Although this is technically correct, to me it seems a bit more logical, and perhaps simpler, to instead use

$$B = \max\{|b_{1}|, |b_{2}|, \ldots, |b_{n_{0} - 1}|, 1 + M\} \tag{2}\label{eq2A}$$

with one less term being used near the end since you already have that with $n = n_0$, you know $|b_{n_0}| \le 1 + M$.