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

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- Apr 5th 2010, 05:55 PMWartonMortonShow 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?

- Apr 5th 2010, 06:21 PMDrexel28
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 $\displaystyle \text{diam }S=\max\text{ }\Omega<\infty$

- Apr 5th 2010, 06:33 PMWartonMorton
- Apr 5th 2010, 06:40 PMDrexel28
- Apr 5th 2010, 06:44 PMWartonMorton
- Apr 5th 2010, 06:48 PMAnonymous1
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...