# analysis

• Nov 24th 2007, 01:39 PM
eigenvector11
analysis
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.
• Nov 24th 2007, 01:53 PM
ThePerfectHacker
Quote:

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.

If $\beta$ is a least upper bound then $\beta - \epsilon$ cannot be an upper bound on the set if $\epsilon > 0$. Why?
• Nov 24th 2007, 02:02 PM
Plato
The statement that $\lambda$ is not an upper bound of S means that $\left( {\exists x \in S} \right)\left[ {\lambda < x} \right]$. That is some number in S is strictly greater that $\lambda$.

The statement that $\beta$ is the least upper bound of S means that $\beta$ is an upper bound of S no number less than $\beta$ is an upper bound.

If $\varepsilon > 0$ then $\beta - \varepsilon < \beta$.

Now you put those three facts together to form a proof of the proposition.