1. Topology proof

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

2. Originally Posted by Danneedshelp
Q: Let A be bounded above so that $s=supA$ exists. Show that $s\in{A^{'}}$, where $A^{'}$ denotes the closure of A.
To say that $s \in A'$, the closure, means that $s \in A$ it is limit point.
If $s\in A$ we are done so suppose that $s\not\in A$, we must show that it is limit point.
By definition of $s=\sup(A)$ if $c>0$ then $s-c$ is not an upper bound.
This $\left( {\forall c > 0} \right)\left( {\exists x_c \in A} \right)\left( {s - c < x_c < s} \right)$
You go to the text material to find out if that is enough to show that it a limit point.

3. Originally Posted by Plato
To say that $s \in A'$, the closure, means that $s \in A$ it is limit point.
If $s\in A$ we are done so suppose that $s\not\in A$, we must show that it is limit point.
By definition of $s=\sup(A)$ if $c>0$ then $s-c$ is not an upper bound.
This $\left( {\forall c > 0} \right)\left( {\exists x_c \in A} \right)\left( {s - c < x_c < s} \right)$
You go to the text material to find out if that is enough to show that it a limit point.
So, since the set $(s-c,s)\cap{A}$ is infinite, we can conclude $s\in{A^{'}}$ for both cases.