Rudin skipped the proof of Step 6.
I want to know if my proof of (M4) is correct. Thanks in advance!
The (M4) in the axioms for multiplication states that
A field $F$ contains an element $1 \neq 0$ such that $1x = x$ for every $x \in F$. Now we are proving that $\mathbb{R}$ has this property.
In Baby Rudin, $\mathbb{R}$ is constructed from $\mathbb{Q}$ as follows:
The members of $\mathbb{R}$ will be certain subsets of $\mathbb{Q}$, called cuts. A cut is, by definition, any set $\alpha \in \mathbb{Q}$ with the following three properties.
(I) $\alpha$ is not empty, and $\alpha \neq \mathbb{Q}$.
(II) If $p \in \alpha$, $q \in \mathbb{Q}$, and $q < p$, then $q \in \alpha$.
(III) If $p \in \alpha$, then $p < r$ for some $r \in \alpha$.
The letters $p$, $q$, $r$, $\dots$ will always denote rational numbers, and $\alpha$, $\beta$, $\gamma$, $\dots$ will denote cuts.
Note that (III) simply says that $\alpha$ has no largest number; (II) implies two facts which will be used freely:
If $p \in \alpha$ and $q \notin \alpha$ then $p < q$.
If $r \notin \alpha$ and $r < s$ then $s \notin \alpha$.
In addition, when proving Step 6, Rudin confine first to $\mathbb{R}^+$, the set of all $\alpha \in \mathbb{R}$ with $\alpha > 0^*$.
If $\alpha \in \mathbb{R}^+$ and $\beta \in \mathbb{R}^+$, we define $\alpha\beta$ to be the set of all $p$ such that $p \leq rs$ for some choice of $r \in \alpha$, $s \in \beta$, $r > 0$, $s > 0$.
We define $1^*$ to be the set of all $q < 1$. It is clear that $1^*$ is a cut.
This is my proof of (M4):
If $r \in \alpha$ and $s \in 1^*$, $r > 0$, $s > 0$, then $rs < r$, hence $rs \in \alpha$. Thus $\alpha \cdot 1^* \subset \alpha$. To obtain the opposite inclusion, pick $p \in \alpha$, and pick $r \in \alpha$, $r > p > 0$. Then $p/r \in 1^*$, and $p = r(p/r) \in \alpha \cdot 1^*$. Thus $\alpha \subset \alpha \cdot 1^*$. We conclude that $\alpha = \alpha \cdot 1^*$.