# Eventually bounded/bounded sequences - I can't complete the proof -_-

• Oct 21st 2012, 05:23 AM
Femto
Eventually bounded/bounded sequences - I can't complete the proof -_-
Proposition

If a sequence is eventually bounded, then it is bounded.

Proof

Suppose $a_n \rightarrow a$ as $n \rightarrow \infty$. Then $\forall \epsilon > 0\ \exists N(\epsilon) \in \mathbb{N}\ s.t.\ |a_n - a| < \epsilon \ \forall n > N$. That is, the subsequence of $a_n$ defined by $a_{N + n}$ is bounded ( $|a_{N + n} - a| < \epsilon$). Since $a_{N + n}$ is bounded $\forall n>N$ it follows that $a_n$ is eventually bounded.

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From here I have no idea where to go. If I'm right, I need to show that as a result of this sequence being eventually bounded, it must be consequently bounded. But I don't see how the first statement implies the second. It says in the margin that I should consider the idea that every set has a maximum and minimum element but at the moment I'm not sure on anything really. Any advice is welcome :) And if this is all completely wrong, please inform me about it.

Also I should note that I have just started my first year at university so a lot of things are a bit alien to me at the moment. Please be patient if I don't understand something that may be above my level :)
• Oct 21st 2012, 08:03 AM
johnsomeone
Re: Eventually bounded/bounded sequences - I can't complete the proof -_-
Your initial proof attempt here begins with the assumption that $a_n$ converges. That's not justified by the assumption that $a_n$ is eventually bounded. For instance, $a_n = (-1)^n$ is certainly eventually bounded, but it doesn't converge.
Try to state what "eventually bounded" means.
• Oct 21st 2012, 08:06 AM
Plato
Re: Eventually bounded/bounded sequences - I can't complete the proof -_-
Quote:

Originally Posted by Femto
Proposition If a sequence is eventually bounded, then it is bounded.
Proof Suppose $a_n \rightarrow a$ as $n \rightarrow \infty$.

Bounded sequences may not converge.
But from the given $\exists B>0~\&~\exists N\in\mathbb{Z}^+$ such that if $n\ge N$ then $\left|a_n\right|\le B$.
Let $B' = \sum\limits_{k = 1}^{N - 1} {\left| {a_k } \right|} + B$ do we have a bound for $a_n~?$
• Oct 21st 2012, 12:49 PM
Femto
Re: Eventually bounded/bounded sequences - I can't complete the proof -_-
Quote:

Originally Posted by Plato
Bounded sequences may not converge.
But from the given $\exists B>0~\&~\exists N\in\mathbb{Z}^+$ such that if $n\ge N$ then $\left|a_n\right|\le B$.
Let $B' = \sum\limits_{k = 1}^{N - 1} {\left| {a_k } \right|} + B$ do we have a bound for $a_n~?$

Thanks.

Yes I realise that bounded sequences may not converge but I thought that all I have to prove is that a sequence is bounded because it is eventually bounded, so I could consider a case in which a sequence is eventually bounded. But yes I see what you mean - that wouldn't be correct.

I've defined $B = max(|a_1|, |a_2|,...,|a_N|, M)$ as my bound for $a_n$. I have seen what you've written before but, and do forgive me if this is a stupid question, what is $B'$ exactly? I see you've defined $B$ to be the bound that exists for all $n \ge N$ but why is $B'$ represented as a sum of bounds? If it is the bound for the whole sequence then shouldn't it be the maximum taken from a set of terms rather than a sum (sorry it just confuses me)?
• Oct 21st 2012, 01:16 PM
Plato
Re: Eventually bounded/bounded sequences - I can't complete the proof -_-
Quote:

Originally Posted by Femto
I've defined $B = max(|a_1|, |a_2|,...,|a_N|, M)$ as my bound for $a_n$. I have seen what you've written before but, and do forgive me if this is a stupid question, what is $B'$ exactly? I see you've defined $B$ to be the bound that exists for all $n \ge N$ but why is $B'$ represented as a sum of bounds? If it is the bound for the whole sequence then shouldn't it be the maximum taken from a set of terms rather than a sum (sorry it just confuses me)?

Both ways work. I just always use the way I did it.
$\forall n$ it is true that $|a_n|\le B'$.