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I wanted to receive some help regarding some homework questions; I appreciate any sort of feedback possible.

I am certain that any Dedekind Cut $A$ has the following properties:

1) $A$ is not the empty set and is NOT the set $\mathbb Q$. 2) For any $a \in A$ and $b$ $\in \mathbb Q$ \ A, $a<b$. 3) $A$ has no maximal element.

Working on defining addition and multiplication for Dedekind Cuts (as the set of all dedekind cuts with the order $\subseteq$ is the set $\mathbb R$.

The additive identity is defined as: $0_\mathbb R$ $:=$ {$y$ $\in \mathbb Q$ such that $y < 0$}.

Question 1: I need to prove that for any Dedekind Cuts $X$ and $Y$, that $X$ + $0_\mathbb R$ $= X$.

This requires me to show that they are both subsets of each other (to be true). By the definition of addition of Dedekind cuts, we have:

$X$ + $0_\mathbb R$ $:=$ {$z$: $z \in \mathbb Q$ such that $z=x+y$, where $x \in X$ and $y \in$ $0_\mathbb R$}.

To show that $X$ + $0_\mathbb R$ $\subseteq X$, we know that for any $z$ in $X$ + $0_\mathbb R$ that $\exists$ $x<a, y<0$ such that $z=x+y$. This implies that $z<a$.

To show that $X \subseteq$ $X$ + $0_\mathbb R$, there exist elements in $0_\mathbb R$ such that they are not contained in $X$. Is this valid?

Question 2: With the multiplication of 2 Dedekind Cuts $X$ and $Y$, I need to show that $XY$ $:=$ { $z$ $\in \mathbb Q:$ there exist $a \in$ $X ∩ \mathbb Q+$, $b \in Y ∩ \mathbb Q+$ such that $z ≤ ab$} is also a Dedekind Cut.

Thus there are three things to show:

a)$\mathbb Q$ \ $XY$ is non empty.

Proof by Contradiction. Assume otherwise, that $\mathbb Q$ \ $XY$ is empty. Then $\mathbb Q \subseteq XY$. Then $∃ z ∈ XY$ such that $z$ is NOT in $\mathbb Q$. Let $z ≤ xy$, $∀ x ∈ X, y ∈ Y$.

Then any $x$ can be represented as $a/b$ and the same with $y$ as $c/d$, where $a,b,c,d ∈ \mathbb Z$, $b,d ≠ 0$.

Then $xy = ab/cd \in \mathbb Q$. But since $z ≤ ac/bd \in \mathbb Q$, then $z ∈ \mathbb Q$ (which is a contradiction).

Thus no such $z$ can exist and so $\mathbb Q$ \ $XY$ is non-empty.

b) Need to show that $∀ z∈ XY$, $∃ z′∈ \mathbb Q$ \ $XY$ such that $z < z′$.

I am having trouble with this part.

c) Need to show that $XY$ has no maximal element.

I am also having trouble with this part.

3) After defining the multiplication of 2 Dedekind Cuts, I am asked to show that the additive inverse, $1_\mathbb R$ exists; that is,

where $1_\mathbb R$ $:=$ {$y \in \mathbb Q$, $y < 0$}.

Prove that $1X=X$.

This requires us to show that $1X ⊆ X$ and vice versa, I assume.

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2 Answers 2

There's too much in this question to answer all at once - all the solutions put together fill up a few pages of notebook paper. I will only do number 1, so you get an idea of how to work with Dedekind cuts. Everything with Dedekind cuts follows straight from the axioms, and you shouldn't really need to resort to proof by contradiction. In the future please try to ask fewer questions in a post.

  1. Let $0_\mathbb{R} = \{y \in \mathbb{Q}|y < 0\}$. Clearly $0_\mathbb{R}$ is a cut. For any two cuts $A$ and $B$, define $X+Y=\{x+y|x \in X, y\in Y\}$. It can be shown that this set is also a cut. Show that for any cut $X$, $X+0_\mathbb{R}=X$.

Proof:

$X + 0_\mathbb{R} \subseteq X$:

Let $z \in X+0_\mathbb{R}$. Then $z = x + y$ for some $x \in X$, $y \in 0_\mathbb{R}$. Since $y < 0$, $z=x+y<x$. Since $z$ is less than an element of $X$, $z \in X$. Since $z$ was arbitrary, $X + 0_\mathbb{R} \subseteq X$.

$X \subseteq X + 0_\mathbb{R}$:

Let $z \in X$. We want to show that $z = x + y$ for some $x \in X$, $y \in 0_\mathbb{R}$. Since $X$ has no maximal element, there exists $x\in X$ such that $z < x$. Let $y=z-x$. Since $z < x$, $y < 0$, so $y \in 0_\mathbb{R}$. Then $z = x + z - x = x + y$, $x \in X$, and $y \in 0_\mathbb{R}$. Thus $z \in X + 0_\mathbb{R}$. Since $z$ was arbitrary, $X \subseteq X + 0_\mathbb{R}$.

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By the way, the answer to this problem in Elementary Analysis Theory of Calculus is wrong. I came here to verify because the answer in the book confused me.

It says,

If r ∈ α and s ∈ 0 ∗ , then r + s < r, so r + s ∈ α. Hence α + 0∗ ⊆ α. Conversely, suppose r ∈ α. Since α has no largest element, there is a rational t ∈ α such that t < r. Then r−t is in 0∗ , so r = t + (r − t) ∈ α + 0∗ . This shows α ⊆ α + 0∗

But, in the book, 0* is defined as above:

0* = {y∈Q|y<0}

So, clearly, the answer meant to say r < t, since t < r => 0 < r-t. So, r-t isn't in 0*. Also, I want to add that it's also good to note that z is in Q, since the set X contains elements of Q. Since the additive property is conservative, x + y for x,y∈Q yields a z that is also in Q. If y or x was a real, then one could not surmise z is in Q, and hence, z may not be in X.

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