Baby Rudin Theorem 1.19 Step 5

Question

I have no background in pure mathematics, and I'm trying to learn how to be more rigorous in general. To help with this, I am trying to make everything more explicit as I progress through Rudin's Principles of Mathematical Analysis. I am currently going through theorem 1.19's proof where Rudin constructs $\mathbb{R}$ from $\mathbb{Q}$.

In particular, in step 5, Rudin states that it is obvious that if $\alpha , \beta , \gamma \in \mathbb{R}$ and $\alpha < \beta$, then $\alpha + \gamma < \beta + \gamma$.

So far, the book has defined $\alpha , \beta , \gamma$ as subsets of $\mathbb{Q}$ called cuts, and has shown that cuts respect the field axioms of addition. I am intuitively convinced that the set $\beta + \gamma$ contains elements that $\alpha + \gamma$ does not, but am struggling with how to formalize it. The following is my thought process on my (failed) attempt.

I know that: $\beta > \alpha \implies \exists x \in \beta : x \notin \alpha$, and I think the next step is the take some value $y \in \gamma$ and show that $y + x \in \beta + \gamma$ while $y + x \notin \alpha + \gamma$. I originally thought this would work by taking $y = \sup \gamma$, but $\sup \gamma$ is a set, not a rational number. When I try to think about taking $y < \sup \gamma$ (edit: sorry, this should say $y \in \gamma$ not $y < \sup \gamma$), it's clear to me that $y + x \in \beta + \gamma$, but it's no longer clear to me that $y + x \notin \alpha + \gamma$. Any help would be greatly appreciated!

Answer 1

If $b\in\beta$ and $b\not\in\alpha,$ then there is some $b'>b,$ such that $b'\in\beta,$ by the definition of a cut.

Let $r=b'-b>0.$ (Intuitively, we have $\alpha+r<\beta.$)

Let $g_1\in \gamma$ and $g_2\not\in\gamma$ such that $g_2-g_1<r.$

We see that $g_1+b'=g_1+b+r\in \beta+\gamma.$

If $g_1+b+r\in\alpha+\gamma,$ then $$g_1+b+r=a+g_3\tag1$$ for some $a\in\alpha,g_3\in\gamma.$

But $a<b$ and $g_3<g_2<g_1+r,$ so:

$$a+g_3<b+g_1+r=b+g_1,$$ which contradicts $(1).$

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We are essentially constructing $g_1,g_2$ as approximations of the real number $\gamma$ on either side of the cut so that $g_2-g_1$ is less than the intuitive real difference $\beta-\alpha.$

What we have is, intuitively, $g_1<\gamma<g_1+r$ and $\alpha <b<b+r<\beta.$

Then for any $a<\alpha,$ and $g<\gamma,$ $a< b$ and $g<g_1+r,$ so $$a+g<b+g_1+r=g_1+(b+r)<\beta+\gamma.$$

This assumes you know that:

Given a cut $\gamma,$ and rational $r>0$ there are rationals $g,h$ with $g\in\gamma$ and $h\not\in\gamma,$ and $h-g<r.$

Answer 2

My personal advice (after many years as a mathematician and teacher of mathematics) is to skip Dedekind cuts and move on. However, to answer your question — I guess you’re not satisfied with Rudin’s argument. Choose $x’<x$ also not in $\alpha$. Choose $y\in\gamma$ so that $y+(x-x’)\notin\gamma$. Then if $x+y\in\alpha+\gamma$, it follows that $x+y=z+w$ for $z\in\alpha$ and $w\in\gamma$. Then $x-z=w-y$, but $x-z>x-x’$. This is a contradiction, as then $w=y+(x-z)>y+(x-x’)$ cannot be in $\gamma$.