# Proof of a Dedekind cut defined by its complement

For a Dedekind cut A:

$-A = \{x \in \mathbb{Q} | -x \in A^c and -x \text{ is not the least element of } A^c\}$

(i) Show that -A is not empty

(ii) Show that -A has no greatest element

This is a worked example from my lecture notes, but I'm having trouble understanding how the proof ties together. I'm fairly comfortable with simple Dedekind cut proofs but this stumps me. It'd be great if someone could help me understand the steps and thinking involved.

The given proof:

(i) Take $y \in A^c$

$\therefore y + 1 \in A^c$

(if $y + 1 \in A \implies y \in A$ which is impossible)

Take $x = -(y + 1)$, $x \in -A$.

(ii) CASE 1: $\exists min(A^c) = a$

$A = (-\infty, a)$ on rationals.

By definition, $-A = \{x \in \mathbb{Q} | -x \ge a \text{and} -x \ne > a\}$

$\therefore -x \gt a$ strictly and so there is no greatest element.

CASE 2: There does not exist min$(A^c)$

This means $-A = \{x \in \mathbb{Q} | -x \in A^c\}$.

If $\exists \text{ max}(-A)$, then $\exists$ min$(A^c)$ = $-\text{max} > (-A)$, which cannot be true.

$\therefore \nexists \text{ max}(-A)$

For (i): By definition $A$ is not $\Bbb{Q}$, so $A^{c}$ is nonempty. We take a $y\in A^{c}$ and to make sure that $y$ is not the least element of $A^{c}$ we look at $y+1$, which is an element of $A^{c}$ because it is closed upwards (as $A$ is closed downwards by definition). Now if you look at the definition of $-A$, you can see that $-(y+1)\in -A$. –  Fanni Jun 26 '13 at 11:46