General Topology, Showing that if $A$ is open then its supremum is not in $A$ unless $A$ is closed Prove: Show that if $A$ is open the sup($A$) is not in $A$ and it is if $A$ is closed.

Definitions: A set $A\in \mathbb{R}$ is open if and only each of its points is an interior point. A point $p\in A$ is a interior point of $A$ if and only if $p$ belongs to some open interval $S_{p}$ which is contained in $A$. A set $A$ is closed if its compliment is open or we could say $A$ is closed if and only if $A$ contains  each of its points of accumulation. A point $p\in \mathbb{R}$ is an accumulation point if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$

proof: Let $A$ be the set of the real numbers. Let $p\in A$ be an interior point of $A$. Then, there exists an open interval $N(p,\epsilon)$ such that $$p\in N(p,\epsilon)\subset A$$ Now, since $A$ is open, then each points of $A$ are an interior point. Now, let $A$ be a partially ordered set $X$ and suppose $a = $ sup($A$)$\in A$, then there exists an $m\in x$ such that $x\leq m$ $\forall x\in A$. Then, since each element of $A$ is an interior point then $a$ also an interior point of $A$. So, there exists a $\delta$ neighborhood of $a$ contained in $A$, so $a + \delta/2\in A$ which contradicts the $a = $ sup($A$). Therefore, $a\notin A$. 

Let's now suppose $A$ is closed, then $p\in \mathbb{R}$ is an accumulation point of $A$ which implies every open set $G$ containing $p$ contains a point of $A$ we will call $q$, where $p\neq q$. 

I am pretty lost with this problem, and I think my reasoning is crap can anyone help me?
I dont think in general it is true that if $A$ is closed then $sup A \in A$ ... for an example set of $\mathbb{N} \subset \mathbb{R}$ but $sup \mathbb{N}$ is not in $\mathbb{N}$ ...if it is given that  $A$ is bounded then that claim for closed set is true...
Sorry for the confusion I have made previously. I have re-written the whole post. Hopefully, it is now much more helpful.

Proposition $1$:
If $E$ is a nonempty set of $\mathbb R$ which is bounded above,
then
(a) sup$E\in \bar E$. In particular, if $E$ is closed in $\mathbb R$, then sup$E\in E$. 

(b) sup$E\notin$ Int $E$. In particular, if $E$ is open in $\mathbb R$, then sup $E\notin E$.

Proof: Let $\alpha$ = sup$E$ which exits in $\mathbb R$ by the least upper bound property of $\mathbb R$. Let $\varepsilon &gt; 0$ be given. There exists $x\in E$ such that $\alpha-\varepsilon &lt; x \leq \alpha$. Hence, every open neighborhood of $\alpha$ intersects with $E$. It follows that $\alpha \in \bar E$ and $\alpha \notin$ Int $E$.

Proposition $2$:
If $E$ is a nonempty set of $\mathbb R$, then sup$E\in \bar E$ (The closure is relative to the set of extended real numbers). In particular, if $E$ is closed in the set of extended real numbers,
then sup$E\in E$.  

Proof:
Case 1: If $E$ is bounded above, then the proof is just the same as Proposition $1$.
Case 2: If $E$ is not bounded above, then sup$E$ = $\infty$. Let $M\in \mathbb R$. Then there exists $x\in E$ such that $M \lt x \lt \infty$. Thus, every open neighborhood of $\infty$ intersects with $E$. It follows that $\infty \in \bar E$.

Comment of the "counterexample" given by the previous reply:


  $\mathbb N \subset \mathbb R$. sup $\mathbb N$ is not in $\mathbb N$ ...


We must be careful about which metric space underlying we refer to. Openness/closeness relative to $\mathbb R$ is not the same as that relative to the set of extended real line.  

$\mathbb N$ is closed relative to $\mathbb R$. However, in the context of allowing the use of extended real numbers as supremum, it makes little sense to consider the induced topology on $\mathbb R$ instead of the topology of the set of extended real numbers. 

$\mathbb N$ is not closed relative to the set of extended real numbers, in which case the extended real number $\infty$ is a limit point of $\mathbb N$. So, $\mathbb N$ is not a counterexample to Proposition 2.