Dedekind's Theorem/Proof 3

Proof of Uniqueness
both $\alpha$ and $\beta$ produce $\tuple {L, R}$.

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 $c < \alpha$.

Because $c < \alpha$, it must be the case that $c \in L$.

Because $\beta < c$, it must be the case that $c \in R$.

That is:
 * $c \in L \cap R$

But by the definition of Dedekind Cut, $\tuple {L, R}$ is a partition of $\R$.

That is, $L$ 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.

Proof of Existence
Let $\gamma$ be the set of all rational numbers $p$ such that $p \in \alpha$ for some $\alpha \in L$.

It is to be verified that $\gamma$ is a cut.

Because $L$ is not empty, neither is $\gamma$.

Suppose $\beta \in \R$ and $q \notin \beta$.

Then because $\alpha < \beta$, we have that $q \notin \alpha$ for any $\alpha \in L$.

Thus $q \notin \gamma$.

Thus $\gamma$ satisfies criterion $(1)$ for being a cut.

Suppose $p \in \gamma$ and $q < p$.

Then $p \in \alpha$ for some $\alpha \in L$.

Hence $q \in \alpha$.

Hence $q \in \gamma$.

Thus $\gamma$ satisfies criterion $(2)$ for being a cut.

Suppose $p \in \gamma$.

Then $p \in \alpha$ for some $\alpha \in L$.

Hence there exists $q > p$ such that $q \in \alpha$.

Hence $q \in \gamma$.

Thus $\gamma$ satisfies criterion $(3)$ for being a cut.

Thus, by definition of the real numbers by identifying them with cuts, $\gamma$ is a real number.

We have that:
 * $\alpha \le \gamma$

for all $\alpha \in L$.

there exists $\beta \in R$ such that $\beta < \gamma$.

Then there exists some rational number $p \in\ Q$ such that $p \in \gamma$ and $p \notin beta$.

But if $p \in \gamma$ then $p \in \alpha$ for some $\alpha\ in L$.

This implies that $\beta < \alpha$.

But this contradicts the definition of Dedekind cut:


 * $(3): \quad \forall x \in L: \forall y \in R: x < y$

Thus $\gamma \le \beta$ for all $\beta \in R$.

That is, $\gamma$ is a producer of $\tuple {L, R}$.