# Dedekind cuts: proof that $A=0_{\mathbb{R}}$ if and only if $-A = 0_{\mathbb{R}}$.

Constructing the real numbers with Dedekind cuts we have the definition of $-A$ for a given $A\in \mathbb{R}$ as

$$-A = \{p\in \mathbb{Q} : -p\in \mathbb{Q}\setminus A, \ \text{and there is} \ q\in \mathbb{Q}\setminus A \ \text{with} \ q < -p \}.$$

We define then $0_{\mathbb{R}}\in \mathbb{R}$ as the cut $0_{\mathbb{R}}= \{q\in \mathbb{Q} : q < 0\}$.

In that case I'm trying to show that $A = 0_{\mathbb{R}}$ if and only if $-A = 0$. For the first part, supposing $A = 0$ we consider $p\in -A$. In that case $-p\in \mathbb{Q}\setminus A$ and since we are supposing $A = 0_{\mathbb{R}}$ this means $-p\geq 0$ so that $p \leq 0$. We want to show that $p\neq 0$ so that $p\in 0_{\mathbb{R}}$. For that suppose $p = 0$, in that case since $p\in -A$ there is $q\in \mathbb{Q}\setminus A$ such that $q < -p$, but then we have $q < 0$ so that $q\in A$ because $A$ is a cut. Since $q\in \mathbb{Q}\setminus A$ we must have $p\neq 0$ and so $p\in 0_{\mathbb{R}}$ showing $-A\subset 0_{\mathbb{R}}$.

Analogously let $q\in 0_{\mathbb{R}}$, so that $q < 0$ and so that $-q > 0$. In that case $q\in \mathbb{Q}\setminus A$. But $0_{\mathbb{R}}$ is a cut, so thare is $r\in 0_{\mathbb{R}}$ with $q < r < 0$. In that case, setting $s = -r$ we have $s > 0$ so that $s\in \mathbb{Q}\setminus A$ and on the same time, $s < -q$. Because of that $q\in -A$ and so $0_{\mathbb{R}}\subset -A$.

With that we have already that $A = 0$ implies $-A = 0$. Now we have to show that $-A = 0$ implies $A = 0$.

I'm stuck with this one however. How can I finish this proof? I've been thinking for a while but couldn't find a good way to do it.

HINT: Show (or use the fact) that $-(-A) = A$ and use the first part that you proved.