Dedekind’s construction of the Reals help I’m stuck interpreting this part of the construction from Rudin’s book:

In Step 4 for (A1) why does taking an $r’ \notin \alpha$ and an $s’ \notin \beta$ mean that their sum is strictly greater than the sum of r and s?
 A: You have two possibilities: either ${r' \in \alpha}$ or ${r' \in \alpha^c}$ (since ${\alpha}$ is a partition of ${\mathbb{Q}}$). It's easy to prove that all rationals in ${\alpha^c}$ are strictly greater than rationals in ${\alpha}$. Likewise ${s' \notin \beta \Rightarrow s' \in \beta^c}$ is greater than every rational in ${\beta}$. This now trivially gives ${r' + s' > r + s}$ for ${r' \in \alpha^c,s' \in \beta^c, r \in \alpha, s \in \beta}$.

How to prove that every element in ${\alpha^c}$ is greater than every element in ${\alpha}$.
So, the cut axioms Rudin uses are:
$$
\begin{array}{ccc}
(1)&\alpha\neq \emptyset,\ \alpha\neq \mathbb{Q}&\\
(2)&p \in \alpha,\ q < p\Rightarrow q \in \alpha&\\
(3)&p \in \alpha\Rightarrow\ \exists\ r \in \alpha\ |\ r > p&
\end{array}
$$
${(2)}$ helps us prove this. Pick a rational ${t \in \alpha^c}$. Then if we had ${t \leq p}$ for some ${p \in \alpha}$, then by ${(2)}$ would tell us ${t \in \alpha}$, which is a contradiction (because ${t}$ cannot be in both ${\alpha}$ and ${\alpha^c}$).
It should also make sense, since you should think of these cuts as basically breaking the real rational line into two sections: a "lower" section, and an "upper" section. So it makes sense all the elements in the upper section are bigger than the elements in the lower section.
Hope that helps.
