Confusion about the proof that if $x \in \mathbb{R}$, $x + (-x) = 0$

Question

On page 19 of Rudin's Principles of Mathematical Analysis, the author constructs real numbers using Dedekind cuts. The proof that the sum of a real number $x$ and its negation $(-x)$ equals $0$ is somewhat clever and more involved. I don't exactly follow it?

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Rudin defines the cut $\beta = -\alpha$ as:

$$\beta := \{p\in\mathbb{Q}:\exists r>0 \text{ such that }(-p-r)\notin \alpha\}$$

In other words, some rational number smaller than $-p$ fails to be in $\alpha$.

We show that $\beta \in \mathbb{R}$ and $\alpha + \beta = 0^{*}$.

If $r \in \alpha$ and $s \in \beta$, then $-s \notin \alpha$. Hence, $r < -s$, so $r + s < 0$. Thus, $\alpha + \beta = 0^{*}$.

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Question. Why does $s \in \beta \implies -s \notin a$ mathemtically?

I mentally picture it as follows:

Answer 1

Suppose for the sake of contradiction $-s\in\alpha$ then for every $r>0$, $-s-r< -s$ and since $\alpha$ is closed downward we would have $-s-r\in\alpha$ for every $r>0$, contradicting the definition of $\beta$.