Bounded above implies there exists a $\sup B$?

Question

This is Rudin's mathematical analysis book's theorem about sup and least-upper-bound

1.11 Theorem

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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

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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?

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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?

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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?

Answer 1

This is the least upper bound property.

Answer 2

The property says

If $S$ is nonempty and bounded above, $\sup S$ exists.

Now Rudin is proving

If $S$ is nonempty and bounded below, $\inf S$ exists.

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 $\sup L=\inf S$?

Answer 3

The fact that if $B\subset S$ is bounded above it has a supremum in $S$ is precisely the least upper bound property for $S$ which is being assumed.

Answer 4

The general result applies to any poset (what is called above ordered set) $P$. If $P$ is such that every nonempty bounded above subset has a supremum, then every nonempty bounded below subset has an infimum. It is also the case that if $P$ is such that every nonempty bounded below subset has an infimum, then every nonempty bounded above subset has a supremum. The proof above is the proof of the former. A very similar proof gives the latter.

In the special case of the real numbers, one either assumed axiomatically that $\mathbb R$ satisfies that every nonempty bounded above subset has a supremum. Or one develops a model of the real numbers from a presupposed model of the rational numbers (typically using Dedekind cuts or Cauchy sequences). One then proceeds to prove that every nonempty bounded above subset has a supremum. In any case, it thus follows that every nonempty bounded below subset has an infimum.

Remark: The general results stated for posets is an example of the principle of duality, which is really what's going on here.