I am reading "Principles of Mathematical Analysis Third Edition" by Walter Rudin.
At p.19 in this book, there is the following statement:
Let $w$ be a positive rational number, and $\alpha$ be a cut. Then, there is an integer $n$ such that $n w \in \alpha$ but $(n + 1) w \notin \alpha$. (Note that this depends on the fact that $\mathbb{Q}$ has the archimedean property!)
The definition of cuts is the following:
The members of $\mathbb{R}$ will be certain subsets of $\mathbb{Q}$, called cuts. A cut is, by definition, any set $\alpha \subset \mathbb{Q}$ with the following properties.
- $\alpha$ is not empty, $\alpha \neq \mathbb{Q}$.
- If $p \in \alpha$, $q \in \mathbb{Q}$, and $q < p$, then $q \in \alpha$.
- If $p \in \alpha$, then $p < r$ for some $r \in \alpha$.
I proved the above statement, but I'm not sure my proof is OK or not.
Is the following proof OK or not?
By definition of a cut, there exist $p$, $q$ such that $p \in \alpha$, $q \notin \alpha$, $q \in \mathbb{Q}$.
By archimedean property of $\mathbb{Q}$, there exists $m \in \mathbb{Z}^{+}$ such that $-p < m w$. So $p > (-m) w$. By the property 2 of cuts, $(-m) w \in \alpha$.
By archimedean property of $\mathbb{Q}$, there exists $n \in \mathbb{Z}^{+}$ such that $q < n w$. By the property 2 of cuts, $n w \notin \alpha$.
So, $S:=\{m \in \mathbb{Z} | m w \in \alpha\}$ is not empty.
So, $T:=\{m \in \mathbb{Z} | m w \notin \alpha\}$ is not empty.
If $m_1 \in S$ and $m_2 \in T$, then $m_1 w < m_2 w$ by the property 2 of cuts.
$\therefore m_1 < m_2$.
$\therefore m_1$ is a lower bound of $T$.
$\therefore$ There exists $\min T$.
Let $n := \min T - 1$.
Then, $(n + 1) w \notin \alpha$, and $n w \in \alpha$.