[Community Question] Calculus: Are there any other books which adopt this axiom of $\mathbb{R}$?

One of our user asked:

I am reading "Calculus" by Takeshi Saito.
In this book, Saito adopts the following axiom of $\mathbb{R}$.
I like this axiom.
I think it is easy to understand what this axiom is saying.

But I cannot find a book in which this axiom of $\mathbb{R}$ is adopted.

Are there any other books which adopt this axiom of $\mathbb{R}$?

Axiom 1.1.1:
1. If $a$ is a real number, then there exists an integer $n$ such that $n \leq a \leq n+1$.
2. If $\{a_n\}$ is a sequence such that $a_i \in \{0, 1\}$ for all $i \in \{1, 2, \cdots\}$, then there exists a real number $b$ such that $$\sum_{n=1}^m \frac{a_n}{2^n} \leq b \leq \sum_{n=1}^m \frac{a_n}{2^n}+\frac{1}{2^m}$$ for all $m \in \{0, 1, 2, \cdots\}$.

By Axiom 1.1.1.1, if $a$ is a real number, there exists a unique integer such that $m \leq a < m+1$ and we define this $m$ as $[a]$. $[a]+1$ is the smallest integer which is greater than $a$.

Proposition 1.1.2:
Let $a, b$ be real numbers.
1. If $a < b$, then there exists $r \in \mathbb{Q}$ such that $a < r < b$.
2. If $|a-b|<\frac{1}{n}$ for all $n \in \{1, 2, \cdots\}$, then $a=b$.

Proof:
1. Let $n:=[\frac{1}{b-a}]+1$. Then, $n$ is the smallest integer such that $n > \frac{1}{b-a} > 0$.
Let $m:=[na]+1$. Then, $na<m\leq na+1 < nb$, so the rational number $r := \frac{m}{n}$ satisfies $a < r < b$.
2. If $|a-b| > 0$, then, by Axiom 1.1.1.1, there exists $n \in \{1,2,\cdots\}$ such that $\frac{1}{n} \leq |a-b| < \frac{1}{n}$. But this is a contradiction. So, $|a-b|=0$. So $a = b$.

Corollary 1.1.3:
If $\{a_n\}$ is a sequence such that $a_i \in \{0, 1\}$ for all $i \in \{1, 2, \cdots\}$, then, there exists a unique real number $b$ such that $$\sum_{n=1}^m \frac{a_n}{2^n} \leq b \leq \sum_{n=1}^m \frac{a_n}{2^n}+\frac{1}{2^m}$$ for all $m \in \{0, 1, 2, \cdots\}$.

Proof:
If $b$ and $c$ satisfy $$\sum_{n=1}^m \frac{a_n}{2^n} \leq b \leq \sum_{n=1}^m \frac{a_n}{2^n}+\frac{1}{2^m}$$ $$\sum_{n=1}^m \frac{a_n}{2^n} \leq c \leq \sum_{n=1}^m \frac{a_n}{2^n}+\frac{1}{2^m}$$ for all $m \in \{0, 1, 2, \cdots\}$, then $|b-c| \leq \frac{1}{2^m} < \frac{1}{m}$. By Proposition 1.1.2.2, $b=c$.

We write this unique real number $b$ as $\sum_{n=1}^{\infty} \frac{a_n}{2^n}$.

Theorem 1.1.4
Let $a, b$ be real numbers such that $a \leq b$.
Let $A$ be a subset of $[a, b]$ which satisfies the following condition (D):

(D) If $x \in A$, then $[a, x] \subset A$.

Then, $A = [a, c]$ or $A = [a, c)$ for some $c \in [a, b]$.

Proof:
First, we show this theorem when $a = 0, b = 1$.
By the condition (D), if $0 \notin A$, then $A = \emptyset = [0, 0)$.
By the condition (D), if $1 \in A$, then $A = [0, 1]$.
So, we assume that $0 \in A$ and $1 \notin A$.

We define a sequence $\{a_n\}$ inductively as follows:
Let $s_0 := 0$
If $s_0 + \frac{1}{2^{0+1}} \in A$, then $a_1 := 1$.
If $s_0 + \frac{1}{2^{0+1}} \notin A$, then $a_1:=0$.
After $a_1, a_2, \cdots, a_m$ are defined, let $s_m := \sum_{n=1}^m \frac{a_n}{2^n}$.
If $s_m + \frac{1}{2^{m+1}} \in A$, then $a_{m+1} := 1$.
If $s_m + \frac{1}{2^{m+1}} \notin A$, then $a_{m+1} := 0$.

By the definition of $\{a_n\}$ and by induction on $m$,
for all $m \in \{0, 1, 2, \cdots \}$, $s_m \in A$ and $s_m + \frac{1}{2^m} \notin A$.

By Axiom 1.1.1.2, there exists a (unique) real number $c := \sum_{n=1}^{\infty} \frac{a_n}{2^n}$ such that $$\sum_{n=1}^m \frac{a_n}{2^n} \leq c \leq \sum_{n=1}^m \frac{a_n}{2^n}+\frac{1}{2^m}$$ for all $m \in \{0, 1, 2, \cdots\}$.

Now we prove that $A = [0, c)$ or $A = [0, c]$.
To prove this, we prove that $[0, c) \subset A$ and $A \cap (c, 1] = \emptyset$.

If $x \in [0, c)$, then, by Proposition 1.1.2.1, there exists an integer $m \in \{1, 2, \cdots \}$ such that $c - x \geq \frac{1}{m}$.
Then, $x \leq c-\frac{1}{m} \leq c-\frac{1}{2^m} \leq s_m$ and $s_m \in A$.
So, by the condition (D), $x \in A$.
$\therefore [0, c) \subset A$.

If $c < x \leq 1$, then by Proposition 1.1.2.1, there exists an integer $m \in \{1, 2, \cdots \}$ such that $x - c \geq \frac{1}{m}$.
Then, $x \geq c+\frac{1}{m} \geq c+\frac{1}{2^m} \geq s_m+\frac{1}{2^m}$ and $s_m+\frac{1}{2^m} \notin A$.
So, by the condition (D), $x \notin A$.
$\therefore A \cap (c, 1] = \emptyset$.

Now we prove the general case.
If $a = b$, then let $c = a = b$.
Then, $A = [a, c]$ or $A = \emptyset = [a, c)$.

If $a < b$, then let $A' := \{\frac{x-a}{b-a} | x \in A \}$.
Then, $A' \subset [0, 1]$ and $A'$ satisfies the condition (D).
Then, there exists $c'$ such that $A' = [0, c']$ or $A' = [0, c')$.
Then, $A = [a, a+(b-a)c']$ or $A = [a, a+(b-a)c')$.