Let $G$ be a Dedekind cut. Show that
$H = \{x \in \mathbb{Q} : $There exists $a \in Q_{>0}$ such that $-x-a \not \in G \}$
Prove that $H$ is a Dedekind cut.
I'm not entirely sure where to begin with this especially since Dedekind cuts are new to me. How would I be able to use the definition of $H$ to prove this?
The Dedekind cut definition is defined as follows: A Dedekind cut $d \subset \mathbb{Q}$ has the following properties:
$d \neq \mathbb{Q}$
$a \in d, b < a \implies b \in d$ (downward closure)
$\forall a \in d, \exists b \in d$ such that $b>a$ (No greatest element)
Any assistance would be much appreciated!
Lets prove first that $H\ne \emptyset$. Assume $H = \emptyset$. That would mean that for all $x \in \Bbb Q$ doesn't exists $a\in\Bbb Q_{>0}$ such that $-x-a\not\in G$. In particular, for all $x\in \Bbb Q$ we have $-x-1 \in G$, so $G = \Bbb Q$, a contradiction. So $H \ne \emptyset$.
Lets prove now that $H \ne \Bbb Q$. Assume that $H = \Bbb Q$. Since $G \ne \emptyset$, choose any $y \in G$. There exist $x \in \Bbb Q = H$ such that $-x < y$, but then exists $a \in \Bbb Q_{>0}$ such that $-x-a \not\in G$, but $-x-a<-x<y \in G$, this contradicts downward closure of $G$.
Can you continue with the next two properties to prove that $H$ is a Dedekind cut?
Hint: The idea is that if $G$ represents the real number $r$, $H$ represents the real number $-r$.