Suppose a set $S$ of real numbers is bounded and let $\mu$ be an upper bound for $S$. Show that $\mu$ is the least upper bound of $S$ $\Longleftrightarrow$ for every $\epsilon > 0$ there is an element of $S$ in the interval $[\mu – \epsilon, \mu]$.
My Work
($\Rightarrow$)
If there is no element of $S$ in the interval $[\mu – \epsilon, \mu]$, then $\mu – \epsilon$ could also be an upper bound for $S$, but since $\mu = \sup S$ and $\mu – \epsilon < \mu$ there is a contradiction.
($\Leftarrow$)
Considering the least upper bound of $S$, I need to show that it cannot be smaller than $\mu$. But I am not sure how to do this using the condition I am given.
Edit
By the definition of a supremum, if $\lambda$ is another upper bound of $S$, then if $\mu = \sup S \Rightarrow \mu \le \lambda$. So proof by contradiction, assuming that $\mu \ne \sup S$, does that mean that there is an element $\lambda \in [\mu – \epsilon, \mu], \lambda \notin S, \lambda < \mu$? How does this prove that there is no element of $S$ in $[\mu – \epsilon, \mu]$ for every $\epsilon > 0$?
Best Answer
Recall the definition of supremum
Now the theorem you might want is
Note the $<$ and not the $\leq$. You will see why this is so.
$(\Rightarrow)$ Suppose $\mu$ is a supremum. Then clearly for any $x\in S$, $x\leq \mu$. Now argue by contradiction. Suppose there were some $\epsilon \;>0$ such that $$\mu-\epsilon \geq x$$ for each element of $A$. This means that $x \leq \mu-\epsilon$ for each $x$. But that would mean $\mu-\epsilon$ is upper bound with $\mu-\epsilon \leq \mu$ which is impossible, since $\mu$ is the supremum.
I'm not sure about the $(\Leftarrow)$. I've always seen this used in one direction. The theorem is sometimes called the approximation theorem, and it says that given a bounded nonempty subset $S$ of the reals, there is a sequence of elements of $S$, call it $s_n$ such that $s_n \to \sup S$,and another sequence, call it $u_n$, such that $u_n \to \inf S$