Dedekind's Theorem/Proof 2

Proof
Let $\alpha = \sup L$.

Let $u_1 \in \R$ such that $u_1 < \alpha$.

$u_1 \in \R$.

Suppose there is no real number $s \in L$ such that $u_1 < s \le \alpha$.

Then $u_1$ would be an upper bound of $L$ which is less than $\sup {L}$.

This is a contradiction.

So there exists at least one real number $s \in L$ such that $u_1 < s \le \alpha$.

But this is also a contradiction because all the elements of $R$ are greater than all the elements of $L$.

It follows that $u_1 \in L$.

Let $u_2 \in \R$ such that $\alpha < u_2$.

$u_2 \in L$.

There exists $s' \in \R$ such that $\alpha < s' < u_2$.

But $s' \notin L$ and thus $s' \in R$.

This is a contradiction because all the elements of $R$ are greater than all the elements of $L$.

It follows that $u_2 \in R$.

Hence, $\alpha$ produces the cut $\tuple {L, R}$.

$\beta \ne \alpha$ also produces $\tuple {L, R}$.

By the Trichotomy Law for Real Numbers either $\beta < \alpha$ or $\alpha < \beta$.

Suppose that $\beta < \alpha$.

From Real Numbers are Close Packed, there exists at least one real number $c$ such that $\beta < c$ and $c < \alpha$.

Then $c \in L \cap R$.

But by the definition of Dedekind Cut, $L$ and $R$ are disjoint.

That is:
 * $L \cap R = \O$

This is a contradiction.

Similarly, $\alpha < \beta$ also leads to a contradiction.

It follows that $\alpha$ is unique.