Real Analysis Proof Help Needed

Let $\displaystyle A$ be bounded below, and define $\displaystyle B = \{b\in\Re : b$ is a lower bound for $\displaystyle A\}$. Show that $\displaystyle \sup B = \inf A$.

This is what I have:

Let $\displaystyle r=\inf A$. Let $\displaystyle b\in B$ be any lower bound for $\displaystyle A$. It follows that $\displaystyle r\geq b$. Since $\displaystyle r\geq $ for all $\displaystyle b \in B$, $\displaystyle r$ is an upper bound for $\displaystyle B$.

Now I need to show that for any other upper bound $\displaystyle s$ for $\displaystyle B$, $\displaystyle r \leq s$. I think this has to do with showing that any upper bound for B is an element of A, but I can't figure out how to show this.

Thanks for the help,

Ultros