# Trouble understanding a part of Rudin's construction of the reals - the additive inverse of a real

I was studying Rudin's Principle of Mathematical Analysis. In his construction of the reals there is a part I am having trouble comprehending. It concerns the additive inverse of the real.

I understand that for a real number $\alpha$ , which is a cut, the additive inverse is given by the cut $\beta$ - which is the set of all rational numbers $p$ with the property that , there exists a rational $r > 0$ such that $-p-r \notin \alpha$ .

Now we need to show that $\alpha + \beta$ is indeed $0$ .

For this we need to show that the cut given by $\alpha + \beta$ is a subset of the cut given by $0$ and the cut given by $0$ is a subset of $\alpha + \beta$ .

I am having trouble understanding proof of the second part viz. the cut given by $0$ is a subset of $\alpha + \beta$ .

As is customary let us denote the cut representing the real $0$ as $0^*$ .

Here is how Ruding approaches the proof :-

Pick a rational $v \in 0^*$ put $w = -v/2$ . Then , $w>0$ and there is an integer $n$ such that $nw \in \alpha$ but $(n+1)w \notin \alpha$ . This is the statement I am having trouble proving rigorously.

Rudin states that this follows from the fact that $\mathbb Q$ has the Archimedean property. Now I appreciate that $\mathbb Q$ has the Archimedean property and that it can be proven without recourse to a least upper bound property which $\mathbb Q$ conspicuously lacks.

But I am unable to prove the existence of $n$ rigorously. I am having trouble doing so because the set of rationals(cut) $\alpha$ does not have a largest element.

Here is how I could somewhat proceed :- Consider the set of integers(let us call it $I$), $m$ such that $mw \in \alpha$ . This set is non-empty . As we know $\alpha$ is non empty and therefore it has at least one rational say $q$ . Now since $\mathbb Q$ has the Archimedean property there exists an integer $m_1$ such that $m_1 w < q$ . Thus the set of integers $I$ is non empty . Similarly we can show that the set $I$ is bounded above. Because $\alpha$ is bounded above. The largest element of $I$ is the $n$ we are looking for. And $(n+1) w$ won't be in $\alpha$.

Is my proof ok ? I am having second thoughts about it.

there is an integer $n$ such that $nw\in \alpha$ but $(n+1)w \notin \alpha$.

The proof is based on the induction axiom (also known as the principle of mathematical induction), which is one of Peano axioms. Rudin does not discuss the Peano axioms because he takes natural numbers as understood.

The induction axiom can be succinctly stated as: every nonempty subset of $\mathbb N$ has the least element. Using the properties of addition/subtraction, we can generalize this to: $\mathbb Z$ has the least upper bound property, as well as the greatest lower bound property. Furthermore, both l.u.b. and g.l.b of a set (if they exist) are contained in the set. Indeed, if $n$ is an upper bound for $A\subset \mathbb Z$ and $n\notin A$, then $n-1$ is also an upper bound for $A$, hence $n$ is not the least upper bound.

Now back to proof. Since $\alpha$ is not all of $\mathbb Q$ there is a rational number $y$ that does not belong to $\alpha$. By the Archimedean property there exists a positive integer $b$ such that $bw>y$. Therefore, $bw\notin \alpha$. In a similar way we can show that there exists an integer $a$ such that $aw\in \alpha$.

Now consider the set $A=\{m\in\mathbb Z: mw\in \alpha\}$. This set is nonempty because it contains $a$. It has an upper bound $b$. Therefore, there exists $n = \sup A$. By the above, $n\in A$. We have $nw\in\alpha$ and $(n+1)w\notin \alpha$, as required. $\Box$

• Thanks. That solves the query. – ameyask86 Jan 10 '13 at 7:11
• "The induction axiom can be succinctly stated as: every nonempty subset of N has the least element" This one really helped. I was implicitly using this fact , but wasn't sure whether to treat it as axiom or not. It is interesting to note that in Apostol's Calculus he gives a proof of the well ordering principle/property of the natural numbers. I guess the property is the same as the induction axion you stated. But Apostol starts with the Least Upper Bound property as an axiom for the reals and uses it to prove the well ordered principle of the natural numbers. Am I right in my understanding? – ameyask86 Jan 10 '13 at 7:19
• @ameyask I'm not familiar with Apostol, but it sounds strange that someone would use lub property of the reals to prove something as basic as an axiom of arithmetics. – user53153 Jan 10 '13 at 7:32
• I guess it has to do with his approach. He states in the beginning of the book that the proper way to go about is to start with construction of rationals from integers and then going on to construct the reals. But since Apostol's book is intended for a much earlier stage in one dealings with mathematics, he takes the properties of real numbers as granted. Thus , Lub is taken as an axiom for the reals. I guess , even in Spivak's calculus it is taken as an axiom earlier on. He then uses it to prove the well ordering principle. – ameyask86 Jan 10 '13 at 10:32
• I guess , if we construct the reals from the rationals , we have to do the other way around. As is done by Rudin. Perhaps Apostol does such a thing because his book is intended more to be an introduction to calculus and a students first exposure to proofs. Thanks again for commenting. – ameyask86 Jan 10 '13 at 10:32