Least Upper Bound Property/Proof 1

Proof
Suppose that $S \subseteq \R_{\ge 0}$ has the positive real number $U$ as an upper bound.

Then $\R_{\ge 0}$ can be represented as a straight line $L$ whose sole endpoint is the point $O$.

Let $l_0 \in \R_{\ge 0}$ be the standard unit of length.

There exists a unique point $X \in L$ such that $U \cdot l_0 = OX$.

Furthermore, if $x \in S$, then:
 * $\map f x = x \cdot l_0$

where $\cdot$ denotes (real) multiplication.

Segments of Finite Lines are Finite
No line segment of $OX$ is infinite.

For suppose that the segment $s$ of $OX$ is infinite.

Then $s$ is greater than every line segment, including any line four times greater than $OX$.

Therefore the less contains the greater: which is impossible.

Existence of Second Endpoint of a Segment of $OX$ beginning at $O$
More precisely, every line segment $s$ of $OX$ having $O$ as one of its endpoints must have another endpoint within $OX$.

The second endpoint of $s$ of $OX$ must exist.

For if the second endpoint does not exist, then $s$ can be continued to any length however great and still remain a segment of $OX$.

But then $s$ can be made over four times as great as $OX$, and still remain a segment of $OX$: which is impossible.

Therefore the second endpoint exists.

Both Endpoints of a Segment of a Line lie Within the Line
Every point of a segment of a straight line $ab$ lies within $ab$.

This second endpoint must be within $OX$.

Formation of the Set $S^*$ Corresponding to the Set $S$
Therefore for every $x \in S$, there is a unique point $\map w x$ such that:
 * $\map f x = O \cdot \map w x = x \cdot l_0$

Thus let $S^*$ be the corresponding set of all respective line segments:
 * $\map f x = O \cdot \map w x$

for all $x \in S$.

We have that $OX$ is greater than or equal to every line segment of $S^*$.

Therefore $OX$ contains every line segment of $S^*$.

And for any two line segments $P, Q$ of $L$ with an endpoint at $O$, either:
 * $P, Q$ are identical
 * $P \subset Q$ but $Q \not \subset P$

or:
 * $Q \subset P$ but $P \not \subset Q$.

Also:
 * $P \subset Q, Q \not \subset P \iff P > Q$

The same can be proven for any other upper bound $OY$ in $S^*$.

Definition of $\Lambda$
Let $\Lambda$ be the union of all line segments of $S^*$.

Existence of $\Lambda$
Let $x \in S$.

Then:
 * $\map f x = O \cdot \map w x$

The point $O$ is an element of $O \cdot \map w x$.

Therefore there is a point $p$ contained in at least one line segment of $S^*$.

But then there must exist an exhaustive and complete figure $F$ containing only all of those points $p$ contained in at least one line segment of $S^*$.

Set theory shows that this figure $F$ is precisely $\Lambda$.

Continuity of $\Lambda$
$\Lambda$ is everywhere continuous.

For, given $p, q \in \Lambda$, such that $p$ and $q$ do not coincide, either $Op > Oq$ or $Op < Oq$.

, let $p$ be less distant from $O$ than $q$.

Then:
 * $\exists x, h \in \R_{\ge 0}: x \in S \land x + h \in S \land p \in O \cdot \map w x \land q \in O \cdot \map w {x + h}$.

But:
 * $O \cdot \map w x \subset O \cdot \map w {x + h}$

Therefore:
 * $p, q \in O \cdot \map w {x + h}$

Therefore $O \cdot \map w {x + h}$ contains every point in between $p, q \in \Lambda$.

Thus suppose $r$ is between $p, q$.

Therefore:
 * $r \in O \cdot \map w {x + h}$

But:
 * $O \cdot \map w {x + h} \in S^*$

Also, $\Lambda$ contains all $p$ in at least one $O \cdot \map w x \in S^*$

Therefore:
 * $r \in O \cdot \map w {x + h}$

Therefore:
 * $\forall p, q \in \Lambda: \forall r: p < r < q: r \in \Lambda$

Therefore $\Lambda$ is everywhere continuous.

$\Lambda$ is Finite
$\Lambda \subseteq OX$.

Let $p \in \Lambda$.

Then:
 * $\exists y \in S: p \in O \cdot \map w y \in S^*$

But it was proven that $OX$ contains every line segment of $S^*$.

Therefore:
 * $p \in O \cdot \map w y \subseteq OX$

Therefore:
 * $p \in OX$

Therefore:
 * $\Lambda \subseteq OX$

Therefore $\Lambda$ has a second endpoint $Z \in OX$, such that $Z$ is between $O, X$.

Therefore $\Lambda = OZ$.

$\Lambda$ is an Upper Bound on $S^*$
$OZ$ is an upper bound on $S^*$. For from set theory it is known that the union of all the sets any given set $D$ contains every set of $D$.

Therefore $OZ$ contains every element of $S^*$.

But then no element of $S^*$ can ever be greater than $OZ$.

Therefore $OZ$ is an upper bound on $S^*$.

$\Lambda$ is the Supremum on $S^*$
$OZ$ is the supremum on $S^*$.

For if $OY$ is any upper bound on $S^*$, $\Lambda \subseteq OY$.

For if $p \in \Lambda$, there is some $y \in S$ such that $p \in O \cdot \map w y \in S^*$.

But it was remarked earlier that if $OY$ is an upper bound on $S^*$, then $OY$ contains every line segment of $S^*$.

Therefore $p \in O \cdot \map w y \subseteq OY$. Therefore $p \in OY$.

Therefore $\Lambda \subseteq OY$.

Therefore $\Lambda \le OY$.

Therefore $OZ$ is an upper bound on $S^*$ and less than or equal to every upper bound on $S^*$.

Therefore $OZ$ is the supremum on $S^*$.

From Supremum is Unique, $OZ$ is unique.

Definition of $z$
There is a unique $z \in \R_{\ge 0}$ such that:
 * $\map f z = O \cdot \map w z = OZ$

$z$ is an Upper Bound on $S$
$z$ is an upper bound on all the elements of $S$.

For if not, then suppose there had been some $x \in S$ such that $x > z$.

$L$ is a representation of the positive real number line.

But if $x \in S$, then $O \cdot \map w x \in S^*$.

Yet $O \cdot \map w x > OZ$, which is impossible because $OZ$ is the supremum on $S^*$.

Therefore $z$ is an upper bound on all the elements of $S$.

$z$ is the Supremum on $S$
$z$ is the supremum on all the elements of $S$. For if $g \in \R_{\ge 0}$ and $g < z$, then $O \cdot \map w g < OZ$. Therefore $O \cdot \map w g$ is not an upper bound on $S^*$. But on the contrary, there exists $\xi \in S$ such that $O \cdot \map w \xi \in S^*$ and $O \cdot \map w \xi > O \cdot \map w g$.

But then $\xi > g$ and $\xi \in S$.

Therefore no $g < z$ can be an upper bound on $S$.

Therefore $z$ is the supremum on all the elements of $S$.

Conclusion
Therefore the set $S \subseteq \R_{\ge 0}$ with upper bound $U$ has the unique supremum $z$.