Let E be a bounded Let $E$ be a bounded subset of $\mathbb{R}$, & let $S$ = sup($E$) be the least upper bound of $E$. $S$ is also a real number. Show that $S$ is an adherent point of $E$, & is also an adherent point of $\mathbb{R}$\ {$E$}.
My Attempt:
For $E$ to be a subset in $\mathbb{R}$, let there exist a $x\in\mathbb{R}$ which is $\epsilon$-adherent to E for every $\epsilon$ > 0. Since $S$ is the sup($E$), since monotone sequences converge, we can say the $S$ = lim($E$) as well. Since $S$ is also the limit of $E$, then it must also be an adherent point of $\mathbb{R}$\ {$E$}. Since $S$ is an adherent point of $\mathbb{R}$\ {$E$}. then it must also be an adherent point of E.
 A: Your attempt looks a little fuzzy... Here is what I would have done :
Since $S = \sup(E)$, and $\forall n \in \mathbb{N}, S-\frac{1}{n}<S$, we have $\forall n \in \mathbb{N}, \exists x_n \in E \cap [S-\frac{1}{n},+\infty[$
Since $S = \sup(E)$, and $\forall n \in \mathbb{N}, S+\frac{1}{n}>S$, if we define $y_n=S+\frac{1}{n}$, we have $\forall n \in \mathbb{N}, y_n \not\in E$
Since $\lim\limits_{n \rightarrow \infty} x_n = S$, $S \in \bar E$
Since $\lim\limits_{n \rightarrow \infty} y_n = S$, $S \in \bar{ \mathbb{R} \setminus E}$
A: Every open Interval $(S-\epsilon,S+\epsilon)$ contains at least one point that is in $E$ and one point that is not in $E$. Otherwise $S$ would not be the least upper bound of $E$:
Because $S$ is an upper bound, the point $S+\epsilon/2$ is not in $E$. On the other hand, if no point in the Interval is in $E$, then $S-\epsilon$ is an upper bound for $E$, contradicting minimality of $S$.
Therefore, every open subset of $\mathbb{R}$ that contains $S$ has nonempty intersection with $E$ and with its complement, so $S$ is an adherent point for both sets.
