Q: Let A be bounded above so that exists. Show that , where 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 exists. Since A is bounded and monotone, there exists a sequence in A such that , by the monotone convergence theorem. To conlude s is a limit point, we need to show that, for any -neighborhood , . However, s may or may not be contained in A, so we must consider both cases. First, if , then given it follows that contains infinitly many points of \ ; 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, , where L denotes the set of all limit points of A (not necessarily in A). Therefore, by definition .