This is Rudin's mathematical analysis book's theorem about sup and least-upper-bound
1.11 Theorem
Suppose $S$ is an ordered set with the least-upper-bound property, $B\subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then
$$\begin{align*}\alpha =\sup L\end{align*}$$
exists in $S$, and $\alpha =\inf B$. In particular, $\inf B$ exists in $S$.
Does this theorem imply this proposition(Is this one theorem or definition?)
proposition
Every subset $B$ of $R$(or ordered sets with least-upper-bound?) which is bounded above must have a $\sup B$?
If not, where is this proposition referred or implied in Rudin's book?
wiki's edition explicitly say: The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.
But in Rudin's book it is constructing Real numbers from Rational numbers. And is on ordered sets firstly?
1.10 Definition
An ordered set $S$ is said to have the least-upper-bound property if the following is true:
If $E\subset S$, $E$ is not empty, and $E$ is bounded above, then $\sup E$ exists in $S$.
If the proposition above is just the Least-Upper-Bound property, then it seems to be an $\color{green}{definition}$, and that is definition 1.10 in Rudin's book. And I found it's an axiom in Royden's Real Analysis.
So, proof: Since $R$ is an ordered set with least upper bound property, then proposition is true. Is this not necessary?