- #1

- 1,462

- 44

## Homework Statement

[/B]

Let ##S\subset \mathbb{R}## be nonempty and bounded above. Show that there must exist a sequence ##\{a_n\}_{n=1}^\infty\subseteq S## such that ##\lim_{n\to\infty}a_n=\sup(S)##.

## Homework Equations

## The Attempt at a Solution

Here is my idea. Let ##\epsilon >0##. Then there must exist an ##a \in S## such that ##\sup(S) - a< \epsilon##, by nature of ##\sup(S)## being the least upper bound.

So clearly we can always find an element ##a## that is arbitrarily close to ##\sup(S)##. My problem is that I don't know how to make this formal. How would I construct the sequence necessary?

What about this: for each ##n \in \mathbb{N}## there must must exist an ##a \in (\sup(S) - 1/n,~ \sup(S))##. And so we construct the sequence out of the ##a##'s that must exist, so that ##a_1 \in (\sup(S) - 1,~ \sup(S))##, ##a_2 \in (\sup(S) - 1/2,~ \sup(S))##, and so on.