Prove that if there is $A\subseteq \mathbb{R}$ not empty and bounded so $infA\leq SupA$ and infA=SupA iff A={One element}.
By definition every bounded set in $\mathbb{R}$ has Inf and Sup, therefore for every $a \in A$ $InfA \leq a \leq SupA\rightarrow InfA\leq SupA$
If A={One element} so A is both Inf and Sup but I can not see why
As you have seen, $a\in A$ implies $\inf A\le a\le \sup A$. Next, if $A$ is not a singleton, say $a,b\in A$ with (wlog.) $a<b$, then $\inf A\le a<b\le \sup A$, i.e., $\inf A<\sup A$. We conclude that $\inf A=\sup A$ can only hold if $A$ is a singleton set. As $A=\{a\}$ implies $\inf A=a=\sup A$, the proof of if and only if is complete.