I am reading "An Introduction to Calculus" by Kunihiko Kodaira.
There is Theorem 1.3 in this book and I am very confused.
We identify a rational number $r$ with a Dedekind cut $(R, R')$ where $R = \{q \in \mathbb{Q} | q < r\}, R' = \{q \in \mathbb{Q} | r \leq q\}$.
In the following proof of Theorem 1.3, $\alpha = (A, A')$ is a Dedekind cut and the elements of $A$ and $A'$ are also Dedekind cuts.
Is it ok?
Are the elements of $R$ and $R'$ Dedekind cuts or original rational numbers?
Assume that $s \in R$ and $s' \in R'$ are Dedekind cuts.
Then, are the elements of $S$ and $S'$ Dedekind cuts or original rational numbers, where $s = (S, S')$?
Theorem 1.3
For any real number $\alpha = (A, A')$,
$$A = \{r \in \mathbb{Q} | r < \alpha \}, A' = \{r \in \mathbb{Q} | r \geq \alpha \}. (1.8)$$
Proof:
To compare a rational number $r$ with a real number $\alpha$, we consider $r$ as a real number and represent it as a Dedekind cut:
$$r = (R, R'), R = \{s \in \mathbb{Q} | s < r\}, R' = \{s \in \mathbb{Q} | s \geq r \}.$$
Since $A \cup A' = \mathbb{Q}, A \cap A' = \emptyset$, it is sufficient to prove that if $r \in A$, then $r < \alpha$ and if $r \in A'$, then $r \geq \alpha$ to prove $(1.8)$.
- Suppose that $r \in A$. If $s < r, s \in \mathbb{Q}$, then $s \in A$. So $R \subset A$. Because $r \notin R$, so $R \neq A$. $\therefore r < \alpha$.
- Suppose that $r \in A'$. If $s \geq r, s \in \mathbb{Q}$, then $s \in A'$. So $R' \subset A'$. So $A \subset R$. $\therefore r \geq \alpha$.
