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I am reviewing what Dedekind cuts are for my quiz tomorrow. I had posted a question before about Dedekind cuts and I thought that was the only problem but there were these two problems as well for this unit I am having trouble understanding. I know what they are and could find it if it was in a real line. but i am having difficulty understanding these 2 problems

For two subsets $X, Y$ of ${\mathbb{O}}$, define the subset $X + Y$ of ${\mathbb{O}}$ by $X + Y = \{x + y |x \in X$ and $y \in Y\}$.

Let $(A_1, A_2)$ and $(B_1, B_2)$ be Dedekind cuts of ${\mathbb{O}}$. Let $C_1 = A_1 + B_1$ (in the above sense) and let $C_2 = {\mathbb{O}} \diagdown C_1$.

Prove that $(C_1, C_2)$ is a Dedekind cut of ${\mathbb{O}}$

and

Let $(A'_1, A'_2)$ be a Dedekind cut of ${\mathbb{O}}$ that represents the same real number as $(A_1, A_2)$. Let $C'_1 = A'_1 + B_1$ and $C'_2 = {\mathbb{O}} \diagdown C'_1$. Prove that $(C'_1, C'_2)$ represents the same real number as $(C_1, C_2)$.

thank you

Ittay Weiss
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1 Answers1

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By Dedekind cut, I take you to mean a pair $(A,B)$ such that:

(i) $A,B\neq\emptyset,$

(ii) $A\cap B=\emptyset,$

(iii) $A\cup B=\Bbb Q,$

(iv) $y\in A$ whenever $y\in\Bbb Q$ and there is some $x\in A$ with $y<x,$ and

(v) $A$ has no greatest element.

(Let me know if that doesn't match your definition.)

Since $A_1,B_1$ non-empty sets of rationals, then so is $C_1$. On the other hand, since $A_2,B_2$ non-empty, then $C_1$ can't be all of $\Bbb Q$, so $C_1,C_2$ partition the rationals. Since $A_1,B_1$ have no greatest element, then neither can $C_1$. Suppose $y\in\Bbb Q$, and that $y<x$ for some $x\in C_1$. Since $C_1=A_1+B_1,$ then $x=a+b$ for some $a\in A_1,b\in B_1$. Take $a'\in\Bbb Q$ such that $a'<a$--so $a'+b<a+b=x$--and $y<a'+b$. (Why can we do this?) Since $(A_1,A_2)$ is a Dedekind cut and $a\in A_1$, then $a'\in A_1$. Put $b'=y-a'$, so that $a'+b'=y<a'+b$, whence $b'<b$, so $b'\in B_1$, and so $y=a'+b'\in C_1$, as desired.

In the second part, they seem to be asking you to show that $A+B$ is well-defined as an operation on left sides of Dedekind cuts. I'm not quite sure what they mean by different Dedekind cuts representing the same real number, though. I was under the impression that each real number corresponded to a unique Dedekind cut.

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Cameron Buie
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  • They may be omitting your condition (v), in which case every rational is represented by two Dedekind cuts. – Brian M. Scott Feb 26 '13 at 01:28
  • Ah! That would make sense. – Cameron Buie Feb 26 '13 at 01:30
  • yes those are the same definitions as mine. actually now it made sense. In my book they had this exactly Take a′∈Q such that a′<a--so a′+b<a+b=x--and y<a′+b. as how you wrote it but i didnt know why that was true. thanks a lot Cameron – MathGeek Feb 26 '13 at 02:48
  • Although for number 2 why do we need to show A + B? arent we trying to show that C1, C2 is same as C'1, C'2? like how you showed in part a can i just implement the same procedure in 2? – MathGeek Feb 26 '13 at 02:50
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    @MathGeek: For the choosing of $a'$, note that $$y<x=a+b,$$ so $$0<a+b-y.$$ Put $$a'=a-\frac{a+b-y}2,$$ which is rational, less than $a$, and $$a'+b=a+b-\frac{a+b-y}2=x-\frac{x-y}2=\frac{x+y}2.$$ Since $x>y$, then $$y<\frac{x+y}2=a'+b,$$ as desired. (cont'd) – Cameron Buie Feb 26 '13 at 04:42
  • (cont'd) As for the second part, my understanding is that two Dedekind cuts represent the same real number precisely when they're the same Dedekind cut, in which case the second part becomes completely trivial. I suspect, therefore, that your definition of Dedekind cut may differ slightly (see Brian's comment for one way that it might differ), or that you're determining which real number is represented by a given cut in a different way. What exactly does it mean (in your book) to say that two Dedekind cuts represent the same real number? – Cameron Buie Feb 26 '13 at 04:46