# Thread: Show that the sequence is bounded.

1. ## Show that the sequence is bounded.

I know that is has to be bounded because its finite, but I'm guessing there is some sort of mathematical proof for this?

2. Originally Posted by WartonMorton
I know that is has to be bounded because its finite, but I'm guessing there is some sort of mathematical proof for this?
Bounded in what sense? That the diameter of the set is finite? This would follow since $\displaystyle \Omega=\left\{|a_k-a_j|:a_k,a_j\in S,\text{ }j\ne k\right\}$ is finite and each element is a finite number and thus $\displaystyle \text{diam }S=\max\text{ }\Omega<\infty$

3. Originally Posted by Drexel28
Bounded in what sense? That the diameter of the set is finite? This would follow since $\displaystyle \Omega=\left\{|a_k-a_j|:a_k,a_j\in S,\text{ }j\ne k\right\}$ is finite and each element is a finite number and thus $\displaystyle \text{diam }S=\max\text{ }\Omega<\infty$

I think that is a little advance. We have not talked about diameters of sequences in class. I think what they mean by bounded is something along the lines of like S= {1, 2, 3, ... , 10} where it is bound by 1 and 10.

4. Originally Posted by WartonMorton
I think that is a little advance. We have not talked about diameters of sequences in class. I think what they mean by bounded is something along the lines of like S= {1, 2, 3, ... , 10} where it is bound by 1 and 10.
Well you still haven't defined what you mean by bounded.

5. Originally Posted by Drexel28
Well you still haven't defined what you mean by bounded.

bounded in the sense of a bounded set..?

6. What Drexel28 is trying to say is...

Assume that $\displaystyle S$ is finite but not bounded above.

Set $\displaystyle max\{S\}=m_0,$ but then the set is bounded by $\displaystyle m_0.$

Therefore, there must be some $\displaystyle m_1\in S: m_1 > m_0,$ but then the set is bounded by $\displaystyle m_1$... ad infinitum.

The set must be infinite to be unbounded.

A similar argument shows the set is bounded below...