Theorem: Least Upper Bound Property

Let $S \subset \R$ be a non-empty subset of the set of real numbers such that $S$ is bounded above.
Then $S$ admits a supremum in $\R$.
This is known as the '''least upper bound property''' of the real numbers.

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.
{{qed|lemma}}
=== 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 $OX$.
{{handwaving}}
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. 
{{qed|lemma}}
=== 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 $OX$.
{{qed|lemma}}
=== 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 $\Lambda$.
{{handwaving|The above needs to be clarified and explained with reference to set-theoretical proofs.}}
{{qed|lemma}}
=== Continuity of $\Lambda$ ===
$\Lambda$ is everywhere continuous.
{{explain|Define "continuous" in this context.}}
For, given $p, q \in \Lambda$, such that $p$ and $q$ do not coincide, either $Op > Oq$ or $Op < Oq$. 
{{WLOG}}, 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. 
{{qed|lemma}}
=== $\Lambda$ 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 $\Lambda = OZ$.
{{qed|lemma}}
=== $\Lambda$ 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 $D$.
{{explain|Link to an appropriate result and express the above sentence in mathematical language.}}
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 $S^*$. 
{{qed|lemma}}
=== $\Lambda$ 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.
{{qed|lemma}}
=== 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 $S$.
{{qed|lemma}}
=== $z$ 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 $S$. 
{{qed|lemma}}
=== Conclusion ===
Therefore the set $S \subseteq \R_{\ge 0}$ with upper bound $U$ has the unique supremum $z$.
{{qed}}