Reals are Isomorphic to Dedekind Cuts

Theorem
Let $\mathscr D$ be set of all Dedekind cuts of the total order $\left({\Q, \leq}\right)$.

Define a mapping $f: \R \to \mathscr D$
 * $\forall x \in \R: f \left({x}\right) = \left\{{y \in \Q: y < x}\right\}$

Then $f$ is a bijection.

Proof
First, we will prove that
 * $\forall x \in \R: f \left({x}\right) \in \mathscr D$

Let $x \in \R$.

It should be proved that $f \left({x}\right)$ is proper subset of $\Q$ such that
 * $(1): \quad \forall z \in f \left({x}\right): \forall y \in \Q: y < z \implies y \in f \left({x}\right)$
 * $(2): \quad \forall z \in f \left({x}\right): \exists y \in f \left({x}\right): x < y$

Because $x \notin f \left({x}\right)$ therfore by definition $f \left({x}\right)$ is a proper subset of $\Q$.

Ad. $(1)$: Let $z \in f \left({x}\right), y \in \Q$ such that
 * $y < z$.

By definition of $f \left({x}\right)$:
 * $z < x$

Then:
 * $y < x$

Thus by definition of $f \left({x}\right)$:
 * $y \in f \left({x}\right)$

Ad. $(2)$: Let $z \in f \left({x}\right)$.

By definition of $f \left({x}\right)$:
 * $z < x$

By Between two Real Numbers exists Rational Number:
 * $\exists r \in \Q: z < r < x$

Then by definition of $f \left({x}\right)$:
 * $r \in f \left({x}\right)$

Thus:
 * $\exists r \in f \left({x}\right): z < r$

By definition of bijection it suffices to prove that $f$ is an injection and a surjection.

We will show by definition that $f: \R \to \mathscr D$ is an injection.

Let $x_1, x_2 \in \R$ such that
 * $f \left({x_1}\right) = f \left({x_2}\right)$

Aiming for a contradiction suppose $x_1 \ne x_2$.

WLOG: assume $x_1 < x_2$.

By Between two Real Numbers exists Rational Number:
 * $\exists r \in \Q: x_1 < r < x_2$

Then by definition of $f \left({x}\right)$:
 * $r \notin f \left({x_1}\right)$

and
 * $r \in f \left({x_2}\right)$

This contradicts $f \left({x_1}\right) = f \left({x_2}\right)$

We will prove by definition thet $f: \R \to \mathscr D$ is a surjection.

Let $L \in \mathscr D$.

By definition of Dedekins cut:
 * $L$ is a proper subset of $\Q$.

By definition of proper subset:
 * $\exists r \in \Q: r \notin L$

By definition of Dedekins cut:
 * $(3): \quad \forall x \in L: \forall y \in \Q: y < x \implies y \in L$

Then
 * $\forall x \in L: r \not< x \land r \ne x$

Hence
 * $\forall x \in L: r > x$

Then $L$ is upper bouded by definition.

By supremum:
 * $\sup\left({L}\right) \leq r$

Hence
 * $\sup\left({L}\right) \in \R$

By definition of supremum
 * $\sup\left({L}\right)$ is upper bound of $L$.

Then by definition of upper bound:
 * $\forall x \in L: x < \sup \left({L}\right)$

We will prove that
 * $\forall x \in \Q: x < \sup\left({L}\right) \implies x \in L$

Let $x \in \Q$ such that
 * $x < \sup \left({L}\right)$

Aiming for a contradiction suppose $x \notin L$

By $(3)$
 * $\forall x \in L: r \geq x$

By definition of upper bound:
 * $r$ is upper bound of $L$.

By definition of supermum:
 * $r \geq \sup \left({L}\right)$

This contradicts $x < \sup \left({L}\right)$

Thus $L = f\left({\sup\left({L}\right)}\right)$