I'm reading Appendix to Chapter 1 in Principles of Mathematical Analysis by Walter Rudin, 3rd edition. In this appendix, Rudin gives a proof of Theorem 1.19 by constructing $\mathbb{R}$ from $\mathbb{Q}$.

Rudin defines the elements of $\mathbb{R}$ to be certain subsets of $\mathbb{Q}$, called cuts, which satisfy these properties:

In the following discussion, $p, q, r, \ldots$ will denote rational numbers, whereas $\alpha, \beta, \gamma, \ldots$ will denote (certain) subsets of $\mathbb{Q}$.

In Step 1, we define a subset $\alpha \subset \mathbb{Q}$ as a cut if $\alpha$ has the following 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 $r \in \alpha$.

In Step 2, we define $\alpha < \beta$ to mean that $\alpha$ is a proper subset of $\beta$, for any two cuts $\alpha$, $\beta$. Then we show that this relation satisfies the trichotomy and the transitivity requirements.

In Step 3, we show that $\mathbb{R}$ has the least upper bound property with respect to the relation $<$ as defined in Step 2.

In Step 4, we define $\alpha + \beta $ to be the set consisting of all the sums $r + s$, where $r \in \alpha$, $s \in \beta$. We show that $\alpha +\beta$ is a cut whenever $\alpha$ and $\beta$ are. The set $0^*$ is defined to be the set of all the negative rational numbers, and this set can be shown also to be a cut. We then show that this operation of addition makes $\mathbb{R}$ into an abelian group with $0^*$ playing the role of $0$. For any $\alpha \in \mathbb{R}$, we define $-\alpha$ as follows: $p \in -\alpha$ if there is some $r > 0$ such that $-p-r \not\in \alpha$. This set $-\alpha$ is also shown to be a cut.

In Step 5, we show that if $\alpha, \beta, \gamma$ are cuts, and if $\beta < \gamma$, then $\alpha + \beta < \alpha + \gamma$. So the relation $<$ has the additive property.

In Step 6, we define $1^*$ to be the set of all the rational numbers less than $1$ and (can) show that this $1^*$ is a cut. We define $\mathbb{R}^+$ to be the set of all the cuts $\alpha$ such that $0^* < \alpha$. Then for any elements $\alpha, \beta \in \mathbb{R}^+$, we define $\alpha \beta$ to be the set of all those elements $p$ such that $p \leq rs$ for some $r \in \alpha$, $s \in \beta$, such that $r > 0$ and $s > 0$. We can show that this set $\alpha \beta $ is also a cut, and we also need to show that the product of two elements of $\mathbb{R}^+$ is in $\mathbb{R}^+$; once we have shown this, then it can be shown that under this operation of multiplication, the set $\mathbb{R}^+$ is a commutative semigroup with $1^*$ as the identity element; moreover, we also show that the operation of multiplication is distributive over addition. Over goal is to show that there is also a multiplicative inverse $1/\alpha \in \mathbb{R}^+$ for each element $\alpha \in \mathbb{R}^+$.

Now my questions are

(1) How do we define $\frac{1}{\alpha}$ for an element $\alpha \in \mathbb{R}^+$?

(2) How to show that if $0^* < \alpha$ and $0^* < \beta$, then $0^* < \alpha \beta$?

For (2), I proceed as follows:

Let $p \in 0^*$. We need to show that $p \leq rs$ for some $r \in \alpha$, $s \in \beta$, such that $r > 0$ and $s > 0$.

Since $p \in 0^*$, we have $p \in \alpha$ and $p \in \beta$. Since $0^* < \alpha$ and $0^* < \beta$, we can find $r \in \alpha$ and $s \in \beta$ such that $r, s \not\in 0^*$. Then using the property (II) of the cuts, we can conclude that $p< r$ and $p< s$.

What next?


(1) You define $\frac{1}{\alpha}$ in a similar way you defined $\alpha\beta$: $\frac{\alpha}{\beta}$ is the set of all rational numbers $q$ such that $q<\frac1r$ for some $r\in \alpha$.

(2) You have $p$ which is a negative number (because $p\in 0^*$) and you have $r$ and $s$ which are nonnegative numbers (because $r\notin 0^*$). The relation $<$ is defined on $\mathbb Q$ and tells you that if $x<y$ and $z>0$, then $xz<xy$.

So, because $0<r$ and $s>0$, you have $0\cdot s < r\cdot s$ or, in other words, $0<rs$.

You also have $p<0$, and because $<$ is transitive, you have $p<rs$

  • $\begingroup$ @xum thanks. But can you please review your answer (1)? I guess you might wanna revise it. $\endgroup$ Jan 24 '16 at 6:23

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