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 $\alpha = (A, A')$,
$$A = \{r \in \mathbb{Q} | r < \alpha \}, A' = \{r \in \mathbb{Q} | r \geq \alpha \}. (1.8)$$

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 $(1.8)$.

  1. 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 $R \neq A$. $\therefore r < \alpha$.
  2. Suppose that $r \in A'$. If $s \geq r, s \in \mathbb{Q}$, then $s \in A'$. So $R' \subset A'$. So $A \subset R$. $\therefore r \geq \alpha$.

A Dedekind cut is always a set of "original" rational numbers, never of other Dedekind cuts. However, in order to consider the rational numbers as a subset of the real numbers, we have to characterise the Dedekind cuts which correspond to rational numbers. That is what they are doing here.


The theorem mentioned in your post is almost trivial. It says that if $\alpha=(A, A') $ is a Dedekind cut then $A$ contains all rationals smaller than $\alpha$ and the rest of rationals (those that are greater than or equal to $\alpha$) lie in $A'$.

The technical issue here is that to compare a rational number with a Dedekind cut requires you to replace the rational number $r$ by the corresponding Dedekind cut $\rho=(R, R') $ where $$R=\{x\in\mathbb {Q} \mid x<r\}, R'=\{x\in\mathbb {Q}\mid x\geq r\} =\mathbb{Q} - R$$ and then compare the cuts $\alpha$ and $\rho$.

By definition of a Dedekind if $\alpha=(A, A') $ and $r\in A$ then there is another rational $s\in A$ with $r<s$. And because of this the cut $\rho=(R, R') $ corresponding to $r$ described in last paragraph will satisfy $\rho<\alpha$. Similarly if $r'\in A'$ and $\rho'$ is the cut corresponding to $r'$ then $\rho'\geq \alpha$.

The theorem is thus an easy/trivial consequence of the definition of Dedekind cut.


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