$S$ is a non-empty subset of $\mathbb{R}$ that is bounded above. Show that the sup$S \in \overline{S}$
My proof: Since $S$ is bounded above, sup$S$ exists in $\mathbb{R}$. Also, by definition, $\overline{S}$ is the smallest closed set of $\mathbb{R}$ containing $S$. So, lets assume that sup$S \notin \overline{S}$. Also, let $k = \text{sup}S$. Now, $k \in \mathbb{R} \setminus{\overline{S}}$ , which is open, as $\overline{S}$ is closed. So there exists open ball $B(k,r) \subset \mathbb{R} \setminus{\overline{S}}$ , i.e., $(k-r, k+r) \subset \mathbb{R} \setminus{\overline{S}}$. But according to definition of sup$S$ , $k- r/2$ is not an upper bound for $S$, thus, $\exists y \in S$ s.t. $y > k -r/2$ , but $y \leq k$ , so $y \in (k-r, k+r)$ , a contradiction. Thus sup$S \in \overline{S}$.
Can some one verify if this is correct, or if it has some loose ends which need to be tied up.
Best Answer
You can also prove this directly (without contradiction), by using the fact that a point is in $\bar S$ if and only if every neighborhood around it contains points of $S$ (prove this using two cases: one case is that the point is in $S$ in which case it is trivial, and the second case is that the point is in the boundary of $S$).
Then, by definition of $s:=\sup S$, for every $\epsilon>0$, there is some $x_\epsilon \in S$ s.t. $|x_\epsilon -s|< \epsilon \iff x_\epsilon \in B(s,\epsilon)$, so indeed $B(s,\epsilon) \cap S \neq \varnothing$ for all $\epsilon>0$ and we are done.