Reals are Isomorphic to Dedekind Cuts

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Theorem

Let $\mathscr D$ be 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:

$(1): \quad \forall z \in \map f x: \forall y \in \Q: y < z \implies y \in \map f x$
$(2): \quad \forall z \in \map f x: \exists y \in \map f x: x < y$

We have that:

$x \notin \map f x$

Therefore by definition $\map f x$ is a proper subset of $\Q$.

Ad. $(1)$: 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 $\map f x$:

$y \in \map f x$

Ad. $(2)$: 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:

$\exists r \in \map f x: z < r$


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:

$L = \map f {\map \sup L}$

$\blacksquare$