Least Upper Bound Property/Proof 1

Theorem
Every class $C \subseteq  \R^+$ having an upper bound  has a least upper bound.

Proof
Suppose that $C \subseteq  \R^+$ has the positive real number $U$ as an upper bound.

Then $\R^+$ can be represented as a straight line $L$ whose sole extremity is the point $O$.

Let $l_0 \in \R^+$ be the standard unit of length.

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

Furthermore if $x \in C$, then $f(x)=x\cdot l_0$.

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 finite straight line, including any line four times greater than $OX$.

Therefore the less contains the greater: which is impossible.

Existence of Second Endpoint of Any 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 degree of 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
This second endpoint must be within $OX$, because every point of a segment of a straight line $ab$ lies within $ab$.

Formation of the Class $C^*$ Corresponding to the Class $C$
Therefore for every $x \in C$, there is a unique point $w(x)$ such that $f(x)= O.w(x)= x\cdot l_0$.

And thus let $C^*$ be the corresponding class of all respective line segments $f(x)= O.w(x)$ for each and every individual $x \in C$.

Then $OX$ contains every line segment of class $C^*$, precisely because $OX$ is greater than or equal to every line segment of class $C^*$.

Any for any two segments $P, Q$ of $L$ beginning at $O$, either: $P, Q$ are identical, or: $P$ contains $Q$ but $Q$ does not contain $P$ or: $Q$ contains $P$ but $P$ does not contain $Q$.

And "$P$ contains $Q$ but $Q$ does not contain $P$" if and only if "$P > Q$".

The same can be proven for any other upper bound $OY$ incumbent upon class $C^*$.

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

Existence of $\Lambda$
For, if $x \in C$, then $f(x)= O.w(x)$.

Then obviously the point $O$ is a member of this line.

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

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 class $C^*$.

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 $p^*$ = whichever one of the two points $p, q$ is less distant from $O$.

Let $q^*$ = whichever one of the two points $p, q$ is more distant from $O$.

Then there exist $x, h \in \R^+$ such that $x \in C \land x+h \in C$ and $p^* \in O.w(x)$ and $q^* \in O.w(x+h)$.

But $O.w(x) \subset O.w(x+h)$.

Therefore $p^*, q^* \in O.w(x+h)$.

Therefore $O.w(x+h)$ contains every point in between $p, q \in \Lambda$.

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

Therefore $r \in O.w(x+h)$.

But $O.w(x+h)\in C^*$.

And $\Lambda$ contains all $p$ in at least one $O.w(x)\in C^*$.

Therefore $r \in O.w(x+h)$.

Therefore for all $p, q$ such that $p,q \in \Lambda$, and all $r$ between $p, q$: $r \in \Lambda$.

Therefore $\Lambda$ is everywhere continuous.

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

For if $p \in \Lambda$, there is some $y \in C$ such that $p \in O.w(y) \in C^*$.

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

Therefore $p \in O.w(y) \subseteq OX$.

Therefore $p \in OX$.

Therfore $\Lambda \subseteq OX$. Therefore $\Lambda$ possess a second extremity $Z \in OX$, such that $Z$ is between $O, X$.

Therefore $\Lambda = OZ$.

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

Therefore $OZ$ contains every member of class $C^*$.

But then no member of class $C^*$ can ever be greater than $OZ$.

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

$\Lambda$ is the Least Upper Bound on $C^*$
$OZ$ is the least upper bound on $C^*$.

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

For if $p \in \Lambda$, there is some $y \in C$ such that $p \in O.w(y) \in C^*$.

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

Therefore $p \in O.w(y) \subseteq OY$. Therefore $p \in OY$.

Therfore $\Lambda \subseteq OY$.

Therefore $\Lambda \le OY$.

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

Therefore $OZ$ is the least upper bound on $C^*$.

Uniqueness of Least Upper Bound
Furthermore, the least upper bound is unique.

Or else we can consider any two least upper bounds on the same class $E$. But one of them would be greater than the other.

But since the greater one is a least upper bound, the lesser one cannot be an upper bound - contrary to the hypothesis!

Definition of $z$
There is a unique $z \in \R^+$ such that $f(z) = O.w(z) = OZ$.

$z$ is an Upper Bound on $C$
$z$ is an upper bound on all the members of class $C$.

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

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

But if $x \in C$, then $O.w(x) \in C^*$.

Yet $O.w(x) > OZ$, which is impossible because $OZ$ is the least upper bound on $C^*$.

Therefore $z$ is an upper bound on all the members of class $C$.

$z$ is the Least Upper Bound on $C$
$z$ is the least upper bound on all the members of class $C$. For if $g\in \R^+$ and $g < z$, then $O.w(g) < OZ$. Therefore $O.w(g)$ is not an upper bound on $C^*$. But on the contrary, there exists $\xi \in C$ such that $O.w(\xi)\in C^*$ and $O.w(\xi)> O.w(g)$.

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

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

Therefore $z$ is the least upper bound on all the members of class $C$.

Conclusion
Therefore the class $C \subseteq  \R^+$ with upper bound $U$ has the unique least upper bound $z$.

Also see

 * Continuum Property implies Well-Ordering Principle