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 $m \in \{0, 1, 2, \cdots\}$.
I will write how to use this axiom to prove an important theorem.
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 $A = [a, a+(b-a)c')$.
This axiom would be proved, ordinarily, by the archimidean property, as J.W.Tanner points out. It's actually not a great "axiom", because it includes things that can be proved from other axioms. Knowing that there's an $n+1$ that's greater than or equal to $x$ -- that's good. There's a least such number by well-ordering. For that $n+1$, we have that $n <= x$. So the left hand side of part 1 is redundant.
Part ii hardly seems to be an axiom at all, for it's easy to prove (assuming you've done the order axioms right) that between any two reals, there's another real, so between the left hand sum, and the LHS plus a little increment, there's a real.
Now you can have, as axioms, things that follow from other axioms -- it's not against the rules or anything. But it seems like a generally bad practice, and not a good approach to use as a model for beginning students.
I think I'd rather stick with the Archimedean principle, and prove the claims of the proposed "axiom" as small lemmas.