Proving, using Dedekind cuts, that $C(0)$ is an additive identity for addition on cuts.

Question

My source is Franck Ayres' Modern Algebra. The author states the fact under discussion (Chapter 7, "Real numbers") but does not provide a proof.

My question is about the second part of the proof, ($\Leftarrow$) direction.

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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 $C$.

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.

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?

Answer 1

The proof that $C \subseteq C + C(0)$ is fairly direct, having nothing to do with $C \subseteq C(0)$ or $C(0) \subseteq C$.

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Suppose $x \in C$. Then [since $C \subset \mathbb{Q}$ is a cut] $x \in \mathbb{Q}$ and there is some $r > x$ such that $r \in C$ (so in particular, $r \in \mathbb{Q}$).

It follows that $x-r < 0$ and $x-r \in \mathbb{Q}$, so $x-r \in C(0)$.

Therefore, $$x = r + (x-r) \in C + C(0)$$

Answer 2

Suppose $ x\in C(r)$, $r\in \mathbb{Q}$. Then $x<r$, and we have that:

$$x<r \implies x -\frac x2 < r - \frac r2 \implies x < r + \frac{x-r}{2}$$

But our hypothesis says that $x-r <0$, so $\frac{x-r}{2} \in C(0)$.