Dedekind's Theorem

Theorem
Let $\left(\mathfrak{U}_1, \mathfrak{U}_2\right)$ be a Dedekind Cut of the set of real numbers $\R$.

Then, there exists exactly one $\alpha \in \mathbb{R}$ such that if $u_1<\alpha$ then $u_1\in \mathfrak{U}_1$ and if $a<u_2$ then $u_2\in\mathfrak{U}_2$. $\alpha$ may be an element of either $\mathfrak{U}_1$ or $\mathfrak{U}_2$.

Thus it is proved that the totally ordered set $\R$ is Dedekind complete, and that is why it is referred to as the continuum.

Proof
Let $\alpha = \sup{\mathfrak{U}_1}$.

Let $u_1\in\mathbb{R}$ and $u_1<\alpha$.

Assume that $u_1\in\mathfrak{U}_2$.

If there are no any real number $s\in\mathfrak{U}_1$ such that $u_1<s\leq\alpha$ then also $u_1$ would be an upper bound of $\mathfrak{U}_1$ which is less than $\sup{\mathfrak{U}_1}$ but this is a contradiction.

Then, there exists at least one real number $s\in\mathfrak{U}_1$ such that $u_1<s\leq\alpha$. But this is a contradiction because all the elements of $\mathfrak{U}_2$ are greater than the elements of $\mathfrak{U}_1$.

Thus, $u_1\in\mathfrak{U}_1$.

Let $u_2\in\mathbb{R}$ and $\alpha<u_2$.

Assume that $u_2\in\mathfrak{U}_1$.

There exists a $s'\in\mathbb{R}$ such that $\alpha<s'<u_2$, but $s'\notin\mathfrak{U}_1$ and thus, $s'\in\mathfrak{U}_2$ and this is a contradiction because all the elements of $\mathfrak{U}_2$ are greater than the elements of $\mathfrak{U}_1$.

Thus, $u_2\in\mathfrak{U}_2$.

Hence, $\alpha$ produces the cut $\left(\mathfrak{U}_1,\mathfrak{U}_2\right)$.

Assume $\beta\neq\alpha$ is also a producer of $\left(\mathfrak{U}_1,\mathfrak{U}_2\right)$.

But either $\beta < \alpha$ or $\alpha < \beta$.

Assume that $\beta < \alpha$, so there exists at least one real number $c$ such that $\beta<c$ and $c<\alpha$. Then $c\in\mathfrak{U}_1\cap\mathfrak{U}_2$ but this is a contradiction because $\mathfrak{U}_1\cap\mathfrak{U}_2=\emptyset$ by the definition of Dedekind Cut. The $\alpha < \beta$ case is similar.

Also see

 * Axiom of Continuity