# If $a_n$ and $b_n$ are equivalent sequences and $a_n$ is bounded then so is $b_n$.

This is what i know;

If $(a_n)$ is an infinite sequence of which is bounded then we can say;

$|a_i| < M$ for all $i \geq 0.$

since $a_n$ and $b_n$ are equivalent sequences, we can say that for every rational $\epsilon >0$, there exists $N \geq 0$ such that

$|a_i-b_i| < \epsilon$ for all $i \geq N$.

I would like to show that $b_n$ is bounded.

now using the triangle inequality we obtain,

$|b_i| = | b_i - a_i + a_i| \leq |b_i - a_i| + |a_i| \leq \epsilon + M$

and hence the sequence is bounded above by $M' = max\{\epsilon, M\}$.

I feel like there is something missing at the end, could someone lend a helping hand please.

• You're definitely on the right path. However, what you've proven is that from some point on, the sequence $b_n$ is bounded by $M+\epsilon$. There might be billions of terms before you get there. You have to point out that all the $b_n$ before that point has a bound as well. – Arthur Jan 24 '15 at 18:35
• If we take $M' = max\{|b_0|,|b_1|,....,|b_{i-1}|, \epsilon + M\}$, then $|b_i| < M'$ for all $i>o$...Is that what you mean? or am i off :|. – user197848 Jan 24 '15 at 18:41
• Yes, that's exactly what I'm talking about. – Arthur Jan 24 '15 at 20:04

Since $a_n$ and $b_n$ are equivalent sequences, we can say that for every rational $1 \geq \epsilon >0$, there exists $N \geq 0$ such that
$|a_i-b_i| < \epsilon$ for all $i \geq N$.
...Then $b_n$ for $n \geq N$ will be bounded by $M' := M+1$.
So $b_n$ is bounded by $\max \{|b_1|,\dots,|b_{N-1}|, M+1\}$
• I like, but whats with the $1 \geq \epsilon > 0$ inequality? i understand that if they are equivilent then this will eventually be true, is there any harm in saying $\epsilon > 0$?. – user197848 Jan 24 '15 at 18:47