Opposite of a Dedekind cut

I'm reading the construction of $\mathbb{R}$ using Dedekind cuts on $\mathbb{Q}$. In my course note, a cut is a nonempty subset $A$ of $\mathbb{Q}$ satisfying:

if $x \in A$ and $y \leq x$ then $y \in A$

Sum of two cuts $A$ and $B$ is defined by:$$A+B = \{x + y, \; x \in A , \, y \in B \}$$The opposition $-A$ of a cut $A$ is defined by$$-A = \{ y, \; y \leq -x \; \forall x \in A \} $$ and the zero cut is $0 = \{ x \leq 0\}$.

Then the note says "it is easy to verify that $A + (-A) = 0$" (on verifie facilement que...) After trying for a while, I still cannot prove this fact:

• in the one hand $A + (-A) \subseteq 0$, indeed if $x \in A$ and $y \in B$ then $y \leq -x$, so $x+y \leq 0$
• in the other hand, how to show that $0 \subseteq A + (-A)$?

Thanks for any hint.