Q: Let A be bounded above so that $\displaystyle s=supA$ exists. Show that $\displaystyle s\in{A^{'}}$, where $\displaystyle A^{'}$ denotes the closure of A.

Below is my try at it. I am not sure about my use of the monotone convergence theorem or the rest of the proof for that matter. We just started basic topology, so I am still getting use to the definitions. Embedded in the proof is a lemma about open intervals and limit points. Not sure if it is needed, because it is not even a very complete proof.

A: Suppose A is bounded above and $\displaystyle s=supA$ exists. Since A is bounded and monotone, there exists a sequence $\displaystyle (a_{n})$ in A such that $\displaystyle (a_{n})\rightarrow\\s$, by the monotone convergence theorem. To conlude s is a limit point, we need to show that, for any $\displaystyle \epsilon$-neighborhood $\displaystyle V_{\epsilon}(s)$, $\displaystyle V_{\epsilon}(s)$$\displaystyle \cap$$\displaystyle A\neq$$\displaystyle \{s\}$. However, s may or may not be contained in A, so we must consider both cases. First, if $\displaystyle s\notin{A}$, then given $\displaystyle \epsilon>0$ it follows that $\displaystyle (s-\epsilon\\,s+\epsilon)$ contains infinitly many points of $\displaystyle A$\$\displaystyle \{s\}$; hence, s is a limit point of A, when A is open. A similar arguement shows that, s is a limit point when A is closed. Moreover, when A is closed it must contain its limit points. Thus, $\displaystyle s\in{A\cup\\L}$, where L denotes the set of all limit points of A (not necessarily in A). Therefore, by definition $\displaystyle s\in{A^{'}}$.