Proof on greatest lower bound and least upper bound. If $A$ is a non-empty bounded subset of $\mathbb{R}$, and $B$ is the set of all upper bounds of $A$, prove that 
$$glb(B)= lub(A)$$
My reasoning and thinking:
The set of upper bounds may be infinite countable set, right? 
 Also all the upper bounds (elements of set $B$) are greater than or equal to $x$ for all $x$ in $A$, using definition of upper bound.
Then it is  $glb(B)$, I am not getting what does it mean? As $y$ is an element of $B$, then is $glb$ of $y$ is $y$ itself?
Any help/hint please.
 A: First, observe that for any $b \in B$, $lub(A) \le b$, 
because $b \in B$ is an upper bound of $A$ and $lub(A)$ is the least of them. So $lub(A)$ is a lower bound of B, hence:
$$lub(A) \le glb(B) \text{.}
$$ 
Because $lub(A)$ is an upper bound of $A$, $lub(A) \in B$ by definition of $B$. Thus, 
$$glb(B) \le lub(A)\text{.}
$$
Hence $lub(A) = glb(B)$.
A: (1)For any  $S\subset R$,  if glb$(S) \in S$ then glb$(S)=\min (S)$.....(2) Let $z=$glb$(B)$. Suppose   $z$ is not an upper bound for $A $ . Then $z<x$ for some $x\in A$. But then (by the definition of $B$), no member of $B$ is less than $x $, so  $x$ is a lower bound for $B$. So $B$ has a lower bound (namely, $x$) that is greater  than the greatest lower bound of $B$ (namely, $z$),which is absurd..... (3.) Therefore  the supposition in (2) is false : So it must be true that $z$ is an upper bound for $A$,that is, glb$(B)=z\in B$. So, by (1), $z$ is the least member of $B $ That is, $z$ is the least of the upper bounds of $A$. $$\text { So we have   lub}(A)=z=\text {glb}(B).$$  
