In baby Rudin we have the definition and theorems:
Definition 1.10 An ordered set $S$ is said to have the $\it{least}$-$\it{upper}$-$\it{bound \hspace{1mm}property}$ if the following is true: If $E\subset S$, $E$ is not empty, and $E$ is bounded above, then $sup(E)$ exists in $S$.
Theorem 1.11 Suppose $S$ is an ordered set with the $\it{least}$-$\it{upper}$-$\it{bound \hspace{1mm}property}$, $B \subset S$, $B$ is not empty, and $B$ bounded below. Let $L$ be the set of all lower bounds of $B$. Then $ \alpha = sup(L) $ exists in $S$, and $\alpha = inf(B)$. In particular, $inf(B)$ exists in $S$.
My question is what is the significance of assuming $S$ has the LUB property? Particularly, since the theorem assumes that $S$ has the LUB property does that imply that there exists another set $E$ (the same one from the definition) in addition to the sets $B$ and $L$-- is the set $L$ the same as the set $E$, or is the set $B$ the same as the set $E$? What confuses me is that the theorem references the set $S$ having the LUB property-- the property being that the set $E$ is bounded $\it{above}$ while in the theorem makes a reference to the set $B$ with almost identical properties to the set $E$ like except the last one, that is $B$ is bounded $\it{below}$. My best guess is that sets $E$ from the definition is the set $L$ in the theorem, can someone confirm or clarify this? Thanks in advance.