A naive question about Dedekind cuts

Question

Let's have a set ${{D}_{l}}\subseteq \mathbb{Q}$. We call it a Dedekind cut (a lower set) if four criteria are met:

1. ${{D}_{l}}\ne \varnothing $;
2. ${{D}_{l}}\ne \mathbb{Q}$;
3. $\forall x,y\left( x\in {{D}_{l}}\wedge y\in \mathbb{Q}\wedge x\ge y\implies y\in {{D}_{l}} \right)$;
4. $\forall x\left( x\in {{D}_{l}}\implies \exists y\left( y\in {{D}_{l}}\wedge x<y \right) \right)$.

We define an upper set ${{D}_{u}}\subseteq \mathbb{Q}$ analogously:

1. ${{D}_{u}}\ne \varnothing $;
2. ${{D}_{u}}\ne \mathbb{Q}$;
3. $\forall x,y\left( x\in {{D}_{u}}\wedge y\in \mathbb{Q}\wedge x\le y\implies y\in {{D}_{u}} \right)$;
4. $\forall x\left( x\in {{D}_{u}}\implies \exists y\left( y\in {{D}_{u}}\wedge x>y \right) \right)$.

Given a set ${{D}_{l}}$, how to formally prove that the set $\mathbb{Q}-{{D}_{l}}$ is definitely NOT a Dedekind cut (an upper set)? Of course, I know that the set ${{D}_{l}}$ doesn't contain the largest element, whereas the set $\mathbb{Q}-{{D}_{l}}$ does contain the smallest element but I'm struggling to prove this obvious fact.

EDIT: Just to clarify, the number represented by the cut is supposed to be rational.

Answer 1

In the set ${\mathbb R}$ of all Dedekind cuts some cuts are special, insofar their lower part is of the form $$D_l:=\bigl\{x\in{\mathbb Q}\bigm| x<a\bigr\}\tag{1}$$ for a certain rational number $a\in{\mathbb Q}$. These special cuts of course give an embedding of ${\mathbb Q}$ into ${\mathbb R}$, and the cuts then are called rational.

When $D_l$ is given by $(1)$ then $a\in{\mathbb Q}\setminus D_l=:D_r$. The number $a$ is in fact the minimal element of $D_r$. Note that according to the order axioms each $y\in{\mathbb Q}$ satisfies exactly one of the relations $$y<a,\quad y=a,\quad y>a\ .$$ By $(1)$ one has $y\in D_l$ iff $y<a$, hence $y\in D_r$ iff $(y=a)\vee(y>a)$.

Answer 2

You cannot prove it, since it's not true.

The complement $\Bbb Q-D_l$ will be an upper cut if and only if the real number it would represent is irrational, i.e. if $D_l$ doesn't have supremum in $\Bbb Q$.

For a specific example, consider $D_l=\{x\in\Bbb Q: x<0\lor x^2<2\}$.