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}}