Reals are Isomorphic to Dedekind Cuts
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Theorem
Let $\mathscr D$ be the set of all Dedekind cuts of the totally ordered set $\struct {\Q, \le}$.
Define a mapping $f: \R \to \mathscr D$ as:
- $\forall x \in \R: \map f x = \set {y \in \Q: y < x}$
Then $f$ is a bijection.
Proof
First, we will prove that:
- $\forall x \in \R: \map f x \in \mathscr D$
Let $x \in \R$.
It is to be proved that $\map f x$ is a proper subset of $\Q$ such that:
- $(1): \quad \forall z \in \map f x: \forall y \in \Q: y < z \implies y \in \map f x$
- $(2): \quad \forall z \in \map f x: \exists y \in \map f x: z < y$
We have that:
- $x \notin \map f x$
Therefore by definition $\map f x$ is a proper subset of $\Q$.
$(1): \quad$ Let $z \in \map f x, y \in \Q$ such that:
- $y < z$
By definition of $\map f x$:
- $z < x$
Then:
- $y < x$
Thus by definition of $\map f x$:
- $y \in \map f x$
$\Box$
$(2): \quad$ Let $z \in \map f x$.
By definition of $\map f x$:
- $z < x$
By Between two Real Numbers exists Rational Number:
- $\exists r \in \Q: z < r < x$
Then by definition of $\map f x$:
- $r \in \map f x$
Thus:
- $\exists r \in \map f x: z < r$
$\Box$
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
- $\map f {x_1} = \map f {x_2}$
Aiming for a contradiction, suppose $x_1 \ne x_2$.
Without loss of generality suppose $x_1 < x_2$.
By Between two Real Numbers exists Rational Number:
- $\exists r \in \Q: x_1 < r < x_2$
Then by definition of $\map f x$:
- $r \notin \map f {x_1}$
and
- $r \in \map f {x_2}$
This contradicts $\map f {x_1} = \map f {x_2}$.
We will prove by definition that $f: \R \to \mathscr D$ is a surjection.
Let $L \in \mathscr D$.
By definition of Dedekind cut:
- $L$ is a proper subset of $\Q$.
By definition of proper subset:
- $\exists r \in \Q: r \notin L$
By definition of Dedekind 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 bounded above by definition.
By definition of supremum:
- $\map \sup L \le r$
Hence:
- $\map \sup L \in \R$
By definition of supremum:
- $\map \sup L$ is an upper bound of $L$.
Then by definition of upper bound:
- $\forall x \in L: x < \map \sup L$
We will prove that:
- $\forall x \in \Q: x < \map \sup L \implies x \in L$
Let $x \in \Q$ such that:
- $x < \map \sup L$
Aiming for a contradiction, suppose $x \notin L$.
By $(3)$:
- $\forall x \in L: r \ge x$
By definition:
- $r$ is an upper bound of $L$.
By definition of supremum:
- $r \ge \map \sup L$
This contradicts $x < \map \sup L$.
Thus:
- $L = \map f {\map \sup L}$
$\blacksquare$