Q : Let A be bounded below and define B$ = \{b \in \mathbb{R} $:b is lower bound for A $\}$ To Show : supB=infA
Proof : Since A is bounded below, lets assume that $a$ is lower bound for A. so $a\in B$. So B is non empty. Also $B=\{ b \in \mathbb{R} : \forall x \in A (b \leq x) \}$. So B is bounded above by A. So B has LUB.
Let $SupB=m$ Claim : m is Lower bound for A If not, then $\exists x \in A (x <m)$. since m=supB. so $\forall \epsilon >0 \exists b \in B (b > m-\epsilon).$Put $ \epsilon = m-x, \exists b \in B (b >x)$, this contradicts the definition of B. So m is lower bound for A.
Let $i$ be other Lower Bound for A. Now $i \in B$. so $i \leq m$. Hence m is infimum of A