How do i show that To show that $\sup B=\inf A$ , where $A$ is set bounded below and B ={$b \in R$ : $b$ is a lower bound for $A$}
Let $x=\sup B$ and $l=\inf A$. Now All lower bounds of A are less than x. so $l \leq x$. How do i proceed to prove tht $l=x$
Set $a=\inf A$. in order to prove that $\sup B=a$ you need to show two things: (a) $b\leq a$ for all $b\in B$ (b) for every $\epsilon>0$, there exist $b\in B$ such that $a-\epsilon<b$.
proof of (a): let $b\in B$ and suppose by contradiction that $a<b$. Set $\epsilon=b-a$. Since $a=\inf A$, there exist $x_0\in A$ such that $x_0<a+\epsilon$, that is $x_0<b$. But by the definition of B, $b\leq x$ for every $x\in A$, so $b\leq x_0$ in particular, which contradicts our assumption.
Proof of (b): Let $\epsilon>0$. Note that $a-\frac{\epsilon}{2}<a\leq x$ for every $x\in A$. Hence, by the definition of B, $a-\frac{\epsilon}{2}\in B$. By setting $b=a-\frac{\epsilon}{2}$ and noticing that $a-\frac{\epsilon}{2}<a-\epsilon$ , we found $b\in B$ such that $a-\epsilon<b$, as required.