Proof that the real number $\text {fin} \mathcal M$ has an additive inverse using induction.

Question

This question regards the proof outlined on page 138 of Fundamentals of Mathematics, Volume 1 Foundations of Mathematics: The Real Number System and Algebra; Edited by H. Behnke, F. Bachmann, K. Fladt, W. Suess and H. Kunle The part giving me trouble is in bold text.

In the following $S\left(\mathcal{M}\right)$ means the set of upper bounds of the set $\mathcal{M}\subseteq\mathbb{Q},$ $\text{fin}\mathcal{M}$ means the least upper bound of $\mathcal{M},$ (AKA, finis superior or supremum) and

$$\mathcal{M}+\mathcal{M}^{\prime}\equiv\left\{x+x^{\prime}\vert{x\in{\mathcal{M}}\land x^{\prime}\in\mathcal{M}^{\prime}}\right\}.$$

We wish to show that the real number $\text{fin}\mathcal{M}$ has an inverse.

For this purpose we consider the set $\mathcal{M}^{\prime}$ of $x^{\prime}$ with $-x^{\prime}\in S\left(\mathcal{M}\right)$ and show that $S\left(0\right)\subseteq S\left(\mathcal{M}+\mathcal{M}^{\prime}\right)$ and also $S\left(\mathcal{M}+\mathcal{M}^{\prime}\right)\subseteq S\left(0\right),$ so that $S\left(0\right)=S\left(\mathcal{M}+\mathcal{M}^{\prime}\right),$ and therefore $\text{fin}\mathcal{M}+\text{fin}\mathcal{M}^{\prime}=0,$ as desired.

For the proof of the first inclusion we assume $z\in S\left(0\right),$ so that $z\ge0.$ From $x\le-x^{\prime}$ for all $x\in\mathcal{M},$$x^{\prime}\in\mathcal{M}^{\prime}$ it follows that $x+x^{\prime}\le z,$ so that $z\in S\left(\mathcal{M}+\mathcal{M}^{\prime}\right).$ Thus we have shown $S\left(0\right)\subseteq S\left(\mathcal{M}+\mathcal{M}^{\prime}\right).$

For the proof of the second inclusion we assume $z\in S\left(\mathcal{M}+\mathcal{M}^{\prime}\right),$ so that $z\ge x+x^{\prime}$ for all $x\in\mathcal{M},$$x\in\mathcal{M}^{\prime},$from which it follows that $z-x^{\prime}\ge x,$ so that $z-x^{\prime}\in S\left(\mathcal{M}\right).$

Since $-x^{\prime}$ is an arbitrary number from $S\left(\mathcal{M}\right),$ we can prove by complete induction that $\left(n+1\right)z-x^{\prime}=z+\left(nz-x^{\prime}\right)$ implies $nz-x^{\prime}\in S\left(\mathcal{M}\right)$ for all natural numbers $n,$ and then for $x\in\mathcal{M}$ we obtain the result that $n\left(-z\right)\le-\left(x+x^{\prime}\right)$ for all $n.$

Since the order is Archimedean, it is impossible that $-z>0.$ Thus $z\ge0,$ and we have completed the proof that $S\left(\mathcal{M}+\mathcal{M}^{\prime}\right)\subseteq S\left(0\right).$

The first part goes as follows: Assume $z\in S\left(0\right):$

$$ z\in S\left(0\right)\implies z\ge0 $$

$$ \implies\forall_{x\in\mathcal{M},x^{\prime}\in\mathcal{M}^{\prime}}\left(x\le-x^{\prime}\implies x+x^{\prime}\le z\right) $$

$$ \implies z\in S\left(\mathcal{M}+\mathcal{M}^{\prime}\right) $$

$$ \implies S\left(0\right)\subseteq S\left(\mathcal{M}+\mathcal{M}^{\prime}\right). $$

I get hung up on the second part. This is what I have so far: Assume $z\in S\left(\mathcal{M}+\mathcal{M}^{\prime}\right):$

$$ z\in S\left(\mathcal{M}+\mathcal{M}^{\prime}\right) $$

$$ \implies\forall_{x\in\mathcal{M},x^{\prime}\in\mathcal{M}^{\prime}}\left(z\ge x+x^{\prime}\implies z-x^{\prime}\ge x\right) $$

$$ \implies z-x^{\prime}\in S\left(\mathcal{M}\right) $$

The second assumption also leads to

$$ \implies\forall_{x\in\mathcal{M},x^{\prime}\in\mathcal{M}^{\prime}}z-x^{\prime}-x\ge0, $$

so

$$ \forall_{x\in\mathcal{M},x^{\prime}\in\mathcal{M}^{\prime},n\in\mathbb{N}_{1}}\left(z-x^{\prime}-x\right)^{n}\ge0 $$

Since $-x^{\prime}\in S\left(\mathcal{M}\right)$ we can choose $-x_{1}^{\prime}\in S\left(\mathcal{M}\right)$ such that

$$ \forall_{x\in\mathcal{M}}\left(-x_{1}^{\prime}-x\right)\ge1. $$

Next we choose $x_{1}$ such that $-x_{1}^{\prime}-x_{1}=1,$ so $\forall_{x\in\mathcal{M}}\left(x_{1}\ge x\right).$ Thus we can find some $-x_{2}^{\prime}\in S\left(\mathcal{M}\right)$ such that $x\le x_{1}<-x_{2}^{\prime}<-x_{1}^{\prime}=1+x_{1}.$

From this we have $z-x_{1}^{\prime}-x_{1}=z+1>z-x_{2}^{\prime}-x_{1}>0.$ Let $\delta=-x_{2}^{\prime}-x_{1}$.

$$ \left(z+1\right)^{n}-\left(z+\delta\right)^{n}>0 $$

$$ \left(1+nz+\dots+z^{n}\right)-\left(\delta^{n}+nz\delta^{n-1}\dots+z^{n}\right)>0 $$

$$ \left(1+nz+\dots\right)-\left(\delta^{n}+nz\delta^{n-1}\dots\right)>0 $$

At some time in the past I learned how to do this very cleanly, but I don't recall how. What is a good way of producing the result indicated in the book?

The way I once knew may have involved setting $z=x^{\prime}+\text{fin}\mathcal{M},$ by an appropriate choice of $x^{\prime}.$

Answer 1

You are right : the proof is quite short and not very clear.

The author choose $z \in S(\mathcal M + \mathcal M')$, so that $z \ge x + x'$ for all $x \in \mathcal M, x' \in \mathcal M'$.

$z$ is a rational whatever, and we want to prove that $z \ge 0$, i.e. that $z \in S(0)$.

From $z \ge x + x'$ it follows that $z — x' \ge x$, so that $z — x' \in S(\mathcal M)$.

What the author has to prove by induction is that :

$nz — x' \in S(\mathcal M)$.

Thus, having proved the base case : $z — x' \in S(\mathcal M)$, where $-x' \in S(\mathcal M)$, we have to prove the induction step :

assuming that $nz — x' \in S(\mathcal M)$, we have to show that :

$(n + 1)z — x' \in S(\mathcal M)$.

The next step of the proof relies on :

$(n +1)z — x' = z + (nz — x')$.

By induction hypotheses : $(nz — x') \in S(\mathcal M)$.

But the base case is : $z + (— x') \in S(\mathcal M)$, for $-x'$ whatever in $S(\mathcal M)$. Thus, we have to consider $(nz — x')$ as the $-x'$ and it's done.

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Having proved $nz -x' \in S(\mathcal M)$, for every $n$, this amounts to : $nz -x' \ge x$, and thus : $nz \ge x+x'$, i.e.

$n(-z) \le - (x+x')$, for every natural number $n$.