Proof involving the sup of a closed and bounded set Prove that if E is bounded and closed then sup E belongs to E
Pf: Let E be bounded and closed so that there $\exists a,b \in R$ such that [a,b]
Let sup E = c
Since c is the least upper bound of E,   $c \le b$
Assume c does not belong to E
Let $\epsilon >0$, 
Therefore $b - \epsilon < c \le b < b+ \epsilon$ but E is bounded so $b +\epsilon$ can't be the least upper bound. Therefore b must be the sup E, which is included in the set E.
Is there a more fluent way to go about this?
 A: Let $A\subseteq\mathbb{R}$ be closed and bounded. Since it is bounded, both inf$(A)$ and sup$(A)$ exist. Let $a=$inf$(A)$ and $b=$sup$(A)$. By definition of sup and inf, for each $\epsilon >0$, $p\in (b-\epsilon,b]$ and $q\in[a,a+\epsilon)$ where $p,q\in A$ and $p\neq b$ and $q\neq a$. Hence, $a,b\in\bar{A}$, and since $A$ is closed, $A=\bar{A}$. Hence, $a,b\in A$.
A: I hope this helps I did this proof in a class I took at Trinity college in Dublin, Ireland.
Show that if $A$ is a bounded set then sup($A$) is an element of $A$ or a point of accumulation.
Definitions/Background Information:
Let $A$ be a partially ordered set $X$. An element $m\in X$ is a lower bound of $A$ if and only if $x\geq M \forall x\in A$. Similarly, an element $M\in x$ is an upper bound of $A$ if and only if $x \leq M \forall x\in A$. If $A$ has both an upper and lower bound, then it is said to be bounded. If some upper bound of $A$ proceeds every other upper bound of $A$, then it is called the least upper bound or supremum of $A$ denoted sup($A$). 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 a bounded set, the sup($A$) is max of $A$ or any $M$, where $M$ is the least element which is greater or equal to $x$ $\forall x\in A$. Let $A\notin A$ or point of accumulation, then select a lower value than $M$ to be the maximum of $A$ which is possible as $M$ not an element nor a point of accumulation. This means $M$ is not the supremum thus contradicting our assumption. Hence $M$ must be an accumulated point or an element of $A$ if M is the supremum.
A: 
Is there a more fluent way to go about this?

Yes (assuming $E$ is a metrizable space, and $E\ne\varnothing$). Let $c=\sup E$. Let $x_n$ be an increasing sequence in $E$, i.e. $x_n\leqslant x_{n+1}$. Then $x_n\leqslant c$ for each $n$, and by definition of $\sup$, for each $n$ there exists $x_n$ such that $c-x_n<\frac1n$. Therefore $c$ is a limit point of $E$, and as $E$ is closed, it follows that $c\in E$.
A: Prove it by contracdiction. First assume that E is closed.Then E contains all its limit points.Let sup E=M.If possible let M is not in E.Then clearly M is not a limit point of E.So there exists d>0 such that N(M,d) contains no element in E.Also since M=sup E.So,for d>0 there exists y belonging to E such that M-d
