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.