If $q \in \alpha$, then $-q \not\in \beta$: Proof by Contradiction. (Theorem 1.19, step 4, (A5), of Mathematical Analysis by Walter Rudin.)

Question

I'm studying the proof for theorem 1.19 from Principles of Mathematical Analysis by Walter Rudin (public online copy here).

The lengthy proof for this theorem is contained in the appendix to chapter 1 on page 17.

I am working on Step 4:

If $\alpha \in R$ and $\beta \in R$ we define $\alpha + \beta$ to be the set of all sums $r + s$, where $r \in \alpha$ and $s \in \beta$.

We define $0^*$ to be the set of all negative rational numbers. It is clear that $0^*$ is a cut. We verify that the axioms for addition (see Definition 1.12) hold in $R$, with $0^*$ playing the role of $0$.

I am specifically working on step 4, (A5), on page 19:

Fix $\alpha \in R$. Let $\beta$ be the set of all $p$ with the following property:

There exists $r > 0$ such that $-p - r \not\in \alpha$.

In other words, some rational number smaller than $-p$ fails to be in $\alpha$.

We show that $\beta \in R$ and that $\alpha + \beta = 0^*$.

For this question, I am focusing on the $\beta \in R$ part.

I encountered the following proposition in Step 4, (A5), on page 19:

If $q \in \alpha$, then $-q \not\in \beta$.

It wasn't immediately clear to me how this was true, and Rudin didn't give an explanation, so I set out to prove it myself. Since we had the keyword "NOT" in the conclusion, I decided that it would be best to use proof by contradiction. The proof is as follows.

A (Hypothesis): $q \in \alpha$, where $\alpha \in R$.

B (Conclusion): $-q \not\in \beta$.

NOT(B): $-q \in \beta$.

I now work forward from the hypothesis and the negation of the conclusion.

NOT(B) 2: There exists $r > 0$ such that $-(-q) - r \not\in \alpha$

$\implies q - r \not\in \alpha$

NOT(B) 3: $q \in Q$ and $r \in Q$, which means that $q - r \in Q$.

A1: $q \in R$, where $R \subset Q$.

NOT(B) 4: $q - r \not\in \alpha \in R \subset Q$, where $q - r \in Q$. But this is a contradiction.

$Q.E.D.$

I would greatly appreciate it if people could please take the take to review my proof for correctness and comment as to whether I went about this correctly.

Answer 1

I think the central issue you are facing is coming to terms with the fact that a real number is defined as a specific kind of set, namely a subset of rationals with some specific properties. You seem to think of a real number as an element of a set (the set $R$ of reals). But set theory axioms do not prevent us to think of elements which are themselves sets like $\{1 \} \in\{\{1 \}, 2\}$.

Next why so many levels of sets are necessary to deal with real numbers? This is something you need to understand carefully. Note that a rational number can be described using two integers, but a generic real number can not be described adequately using a finite number of rationals. Thus for example what do you mean by $\sqrt{2}$ and how do you compare it with existing rationals? You say that $\sqrt{2}$ is some positive number $x$ such that $x^2=2$. This helps somewhat but does not give a complete picture. How do you compare it with other numbers? Thus can you say it is greater than $1$? Yes! Can you say it is greater than $1.4$? Yes! But so what, there are many numbers greater than both $1, 1.4$ (for example $1.5$). So you need more information to describe (at least for comparison with existing rational numbers) the real number $\sqrt{2}$. More precisely you want it to be positive and greater than all positive rationals whose squares are less than 2 and also less than all positive rationals whose squares are greater than 2. In symbols if $x, y$ are positive rational numbers with $x^2<2<y^2$ then you want $x<\sqrt{2}<y$. A complete picture of $\sqrt{2}$ thus requires you to deal with all such rationals numbers $x$.

Thus we define the symbol $\sqrt{2}$ to be a set of all such positive rationals $x$ with $x^2<2$. Some people might try to define it as set of all $y$ with $y^2>2$. Dedekind in fact used both the sets of $x$ and $y$ together. And each such way of defining real numbers is valid.

To proceed with the notion of a general real number, we have to partition the whole set $Q$ of rational numbers into two sets $L, U$ such that $\emptyset\neq L\neq Q$ and $U=Q-L$ and we further postulate another property of these sets: if $x\in L$ then for any rational $y<x$ we should have $y\in L$ (this automatically implies that if $x\in U$ and $y>x$ then $y\in U$. Check this! Dedekind on the other hand expressed this by postulating that if $x\in L$ and $y\in U$ then $x<y$). Dedekind said that a real number is just such a pair of sets $L, U$. Rudin says that a real number is just the set $L$.

When the set of rationals is partitioned into two sets $L, U$ described in the manner above then there are three mutually exclusive and exhaustive possibilities :

1. Either $L$ has a greatest member $l$.
2. Or $U$ has a least member $u$.
3. Neither $L$ has a greatest member nor $U$ has a least member.

In both the cases 1) and 2) above we say that the real number being defined using these sets is a rational real number which is $l$ in 1) and $u$ in 2). Case 3) defines an irrational real number. Thus when we try to represent a rational number via this mechanism we have two options and for convenience we choose one option namely case 2) and thus while defining a Dedekind cut we postulate another property that $L$ does not have a greatest member.

Rudin's theorem 1.19 describes this procedure with heavy amount of symbolism/formalism and also describes how to perform arithmetic operations with these Dedekind cuts.