# Math Help - analysis homework

1. ## analysis homework

Let S be a nonempty set of real numbers that is bounded above, and let Beta be the least upper bound of S. Prove that for every Epsilon greater than 0, there exists an element x such that x is greater than Beta minus Epsilon.

Any ideas or help would be appreciated. Thanks.

2. Originally Posted by eigenvector11
Let S be a nonempty set of real numbers that is bounded above, and let Beta be the least upper bound of S. Prove that for every Epsilon greater than 0, there exists an element x such that x is greater than Beta minus Epsilon.

Any ideas or help would be appreciated. Thanks.
note that $\beta - \epsilon < \beta - \frac {\epsilon}2 < \beta$

3. I'm not sure where to go with your little hint. Are you saying that Beta minus (epsilon/2) will be my x value?

4. Originally Posted by eigenvector11
I'm not sure where to go with your little hint. Are you saying that Beta minus (epsilon/2) will be my x value?
Yes.

-Dan

5. Originally Posted by eigenvector11
I'm not sure where to go with your little hint. Are you saying that Beta minus (epsilon/2) will be my x value?
yes

EDIT: haha, jinx, topsquark! you owe me a soda!

6. Originally Posted by eigenvector11
I'm not sure where to go with your little hint. Are you saying that Beta minus (epsilon/2) will be my x value?
Sorry but that argument does not work.
How do you know that $\left( {\beta - \frac{\varepsilon }{2}} \right) \in S$? The fact is you do not know that.

What does work is quite simple. What does least mean?
If $\beta$ is least then $\beta - \varepsilon$ is not.

7. But how do you know that there is another element other than Beta that is greater than (Beta - Epsilon)?

8. Originally Posted by Plato
Sorry but that argument does not work.
How do you know that $\left( {\beta - \frac{\varepsilon }{2}} \right) \in S$? The fact is you do not know that.
ah yes. i was thinking of S as an interval, but that might not be true

9. Does anyone else have any ideas for this question? My assignment is due fairly soon. Thanks.