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?
Best Answer
The property says
Now Rudin is proving
He does this by noticing that the set $L$ of lower bounds of $S$ is nonempty and bounded above, so $\sup L$ exists. Can you see why $\sup L=\inf S$?