1. ## Topology proof

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 $\displaystyle s\in{A^{'}}$.

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