My source is Franck Ayres' Modern Algebra. The author states the fact under discussion ( Chapter $7$Chapter 7, " Real"Real numbers") butbut does not provide a proof.
My question is about the second part of the proof, ( $<==$$\Leftarrow$) direction.
$\begin{align*} &\text{Let } C \text{ be a Dedekind cut. } C \text{ can be written as } C(r) = \{ x \mid x < r \text{ for some } r \in \mathbb{Q} \}. \\ &\text{The goal is to prove that: } C + C(0) = C. \text{ I first prove that } C + C(0) \text{ is included in } C. \\ &\text{Suppose } x \text{ belongs to } C + C(0) = \{ c + c_0 \mid c \in C \text{ and } c_0 \in C(0) \}. \end{align*}$ Let $C$ be a Dedekind cut. $C$ can be written as $C(r) = \{ x \mid x < r \text{ for some } r \in \mathbb{Q} \}$. The goal is to prove that: $C + C(0) = C$. I first prove that $C + C(0)$ is included in $C$.
$\begin{align*} &\text{Because } x \text{ belongs to } C + C(0), x = c + c_0 \text{ for some } c \in C \text{ and some } c_0 \in C(0). \\ &\text{We have: } c < r \text{ and } c_0 < 0. \\ &\text{Therefore, } c+c_0 < r + 0 = r \text{ implying that } x \in C. \\ &\text{So, for all } x, x \in C + C(0) \text{ implies } x \in C. \end{align*}$ Suppose $x$ belongs to $C + C(0) = \{ c + c_0 \mid c \in C \text{ and } c_0 \in C(0) \}$.
Because $x$ belongs to $C + C(0)$, $x = c + c_0$ for some $c \in C$ and some $c_0 \in C(0)$.
We have: $c < r$ and $c_0 < 0$. Therefore, $c+c_0 < r + 0 = r$ implying that $x \in C$. So, for all $x$, $x \in C + C(0)$ implies $x \in C$.
Second part of the proof : $<==$$\Leftarrow$ direction.
$ \begin{align*} &\text{I now want to show that: } C \subseteq C+C(0). \\ &\text{I first consider the case in which } C \text{ is a negative cut, meaning that } C \subseteq C(0). \\ &\text{Suppose that } x \in C. \\ &\text{Because multiplication by a positive rational preserves order, } x < r \text{ implies } x/4 < r. \\ &\text{Set } c = x/4, \text{ we have } c < r \text{ and } c \in C. \\ &\text{Also, set } c_0 = c = x/4. \\ &\text{This implies } c_0 \in C \text{ and consequently } c_0 \in C(0). \\ &\text{It follows that } c+c_0= 2c = 2(2(x/4))=x \\ &\text{ meaning that } x \text{ can be written as the sum of two elements of } &\\C &\text{ and } C(0) \text{ as desired.} &\end{align*} $ I now want to show that: $C \subseteq C+C(0)$. I first consider the case in which $C$ is a negative cut, meaning that $C \subseteq C(0)$.
Suppose that $x \in C$. Because multiplication by a positive rational preserves order, $x < r$ implies $x/4 < r$. Set $c = x/4$, we have $c < r$ and $c \in C$. Also, set $c_0 = c = x/4$. This implies $c_0 \in C$ and consequently $c_0 \in C(0)$. It follows that $$c+c_0= 2c = 2(2(x/4))=x,$$ meaning that $x$ can be written as the sum of two elements of $C$ and $C(0)$ as desired.
My problem is that I cannot find a trick to write $x$ as $ x = c+c_0$ ( with $c\in C $ and $c_0\in C(0)$) in the second case, namely, when $C$ is not included in $C(0)$, but, the other way around, when $C(0)\subseteq C$.
Or, does the problem lie in my decision to distinguish 2 cases?
