closed subset of $\Bbb R$ shows least upper bound property Can we prove without using the least upper bound property in $\Bbb R$ that any closed subset $E$ of $R$ which is bounded above has a least upper bound in $\Bbb R$. If a subset of $\Bbb R$ contains all its limit point then the highest limit point is the least upper bound of that subset. But, how to write it formally ?  
 A: It depends on what you’re allowed to assume about $\Bbb R$; here’s one possible argument that doesn’t use the least upper bound property directly.
Let $A$ be a non-empty closed subset of $\Bbb R$, and assume that $b$ is an upper bound for $A$. If $A$ has a maximum element, we’re done, so assume that it does not. Let $A'=\bigcup_{a\in A}(\leftarrow,a]$, and let $B$ be the set of all upper bounds for $A$. If $x\in A'$ and $y\in B$, then there is an $a\in A$ such that $x\le a\le y$, so $x\le y$. Now let $x\in\Bbb R\setminus A'$; then for each $a\in A$ we have $x\not\le a$, so $a<x$, and $x\in B$. Thus, $A'\cup B=\Bbb R$.
Let $C=A'\cap B$. If $x,y\in C$ with $x<y$, then there is an $a\in A$ such that $x<y\le a$, and hence $x\notin B\supseteq C$, so $|C|\le 1$. If $C=\{x\}$, it’s not hard to show that $x=\max A=\min B$. Thus, $C=\varnothing$, since we assumed that $A$ has no maximum element. If $B$ has no minimum element, then $\{A',B\}$ is a disconnection of $\Bbb R$, which is impossible, so let $x=\min B$. Then for each $y<x$ there is an $a_y\in A$ such that $y<a_y<x$, so $x\in\operatorname{cl}A=A$, a contradiction that establishes the desired result.
A: Let a be that highest limit point and B be any upper bound. Then a is an upper bound of the closed subset.  We have that a must be ≤ B because if B < a then B is not an upper bound.  Since this is true for any upper bound B, then a has to be the least upper bound. 
The only point to pin down is that there is an "a" which is the highest limit point.  That is ensured by the fact that there is an upper bound on the subset.  That is, no limit point a can be > B; and the subset is closed so a must be within it. 
