Prove that $\text{glb}(A)=\text{lub}(A) \iff A$ contains just one element. 
Prove that $\text{glb}(A)=\text{lub}(A) \iff  A$ contains just one element.

I understand that if a set has only one element, say $x$, then $x$ would be the lower upper bound as well as the greatest lower bound. I am trying to figure out what laws or theorems I need to use in order to back this statement up with.
 A: Let $A = \{x\} \subset \mathbb{R}$.  $A$ is bounded above by $x$, since $\forall y \in A, y \le x$.  Suppose $\exists \delta \in \mathbb{R}$ such that $\delta \lt x$ and is an upper bound of $A$.  Then $\forall y \in A, y \le \delta$.  But  $x \in A$, so $x \le \delta$, a contradiction. Therefore , $x = sup A$.
In a similar manner it can be shown that $x= inf A$.  Thus $sup \{x\} = inf \{x\}$, as desired. 
Let $B\subset\mathbb{R}$.  Assume $b = sup B = inf B$.  Then $\forall x\in B, inf B \le x \le sup B$.  But then $\forall x\in B, b\le x \le b$. This can only hold if $x=b, \forall x\in B$.  Therefore, $B = \{b\}$ .
A: If $A=\{a\}$ then $\text{glb} A=\text{lub} A=a$
Conversely assume that $A$ has more than one element.Then order of the set $A$ is atleast $2$.
Let $\{p,q\}\subset A;$. Then by the usual ordering of real numbers either $p<q$ or $q<p$.
If $p<q$ then obviously $\text{glb}(A)\le p$ and $\text{lub} (A)\ge q$ and hence $\text {glb} (A)\neq \text{lub} (A).$ contradiction.
Same reasoning for $q<p$.
