Consider: Let $y \in \mathbb{R}. Q_y = \{q \in \mathbb{Q} \mid q < y\}$. I would like to prove that $Q_y$ is a Dedekind cut, following the definitions below.
$d \neq \mathbb{Q}$ and $d \neq \emptyset$
$a \in d, b \in \mathbb{Q}, b < a \implies b \in d$ (downward closure)
$\forall a \in d, \exists b \in d$ such that $b>a$ (No greatest element)
I'm having trouble showing 2. Parts 1 and 3 seem straight forward since $Q_y \neq \mathbb{Q}$ and $Q_y \neq \emptyset$ where $y \in \mathbb{R}$. Part 3 also follows from $\sup(Q_y) = y$. I'm not convinced though that $Q_y$ has downward closure. The more I try to formulate a proof the less I think it's true, but I know that $Q_y$ is certainly a Dedekind cut.
Could someone please help with the proof of property 2 for this question?
It’s pretty easy, you want to show that if $x\in Q_{y}$ and $z<x$ then $z\in Q_{y}$ but this is true because: if $x\in Q_{y}$ then $x<y$, so if $z<x$ then we must have $z<y$ and so $z\in Q_{y}$.Hence 2 is true.