Let A be a bounded subset of R "REAL NUMBERS" B a nonempty subset of A .show that :
$\displaystyle inf A \leq inf B \leq supB \leq sup A $
For all $\displaystyle x\in B,$ $\displaystyle x\in A$ (since $\displaystyle B\subseteq A)$ and so $\displaystyle x\leqslant\sup A$ (as $\displaystyle \sup A$ is an upper bound for $\displaystyle A).$ Hence $\displaystyle \sup A$ is an upper bound for $\displaystyle B$ and so $\displaystyle \sup B\leqslant\sup A$ (since $\displaystyle \sup B$ is the least of the upper bounds for $\displaystyle B).$
A similar argument shows that $\displaystyle \inf A\leqslant\inf B.$
Finally, as $\displaystyle B\ne\O,$ let $\displaystyle b\in B.$ Then $\displaystyle \inf B\leqslant b\leqslant\sup B$ (since $\displaystyle \inf B$ and $\displaystyle \sup B$ are a lower and an upper bound for $\displaystyle B$ respectively).