
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 28th, 2014, 01:30 PM  #1 
Member Joined: May 2014 From: None of your business Posts: 31 Thanks: 0  Addition with Dedekind cuts, proof
I'm trying to show that $r +_R s$ is a real number, i.e. a Dedekind left set. With this notation $r$ and $s$ are both Dedekind left sets. I'm using the definition that $r+_Rs$ = {$p+_Qq : p \in r$ and $q \in s$} In particular, I'm having some difficulty showing $r +_R s$ is closed to the left, i.e. if $q \in {r+_Rs}$ and $p <_Q q$, then $p\in {r+_Rs}$. What I have done thus far is... Suppose $a+_Qb \in r+_Rs$. Further suppose that $c,d \in Q$ and $c+_Qd<_Qa+_Qb$. (*) Then there is some $j \in r$ and $k \in s$ such that $j+_Qk=c+_Qd$. Since $j \in r$ and $k \in s$, then $j+_Qk$, which is equal to $c+_Qd$, must be in $r+_Rs$. (I feel like I'm stretching the concept of equality a bit here). Now I made the statement (*), because I thought of a counterexample to the statement: Then $c \in r$ and $d \in s$. Here's my counterexample: Let $c=7$ and $d=8$. Then $7+_Q8=1$ and $1 \in 3+_R4$, but $ 8 \ni 4$ and $8 \ni 3$. Is the statement (*) legit? It seems very obvious to me, but I'm having a hard time proving it. Please help. This is more of a set theory question, but since it involved the construction of the reals, I thought it best to place this thread in the real analysis section. 
June 28th, 2014, 09:18 PM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
The problem, as you probably noticed, happens when you add a negative number to a positive one. This makes being precise "conceptually difficult". So let's just start with: $p <_{\Bbb Q} q$. Now $q \in r +_{\Bbb R} s$, so that means: $q = a +_{\Bbb Q} b$, for $a \in r, b \in s$. Since $q = p +_{\Bbb Q} (q _{\Bbb Q} p) = a +_{\Bbb Q} b$ $p = (a + (p _{\Bbb Q} q)) +_{\Bbb Q} b$. So we just need to show that $a + (p _{\Bbb Q} q) \in r$. But this follows because $a + (p _{\Bbb Q} q) <_{\Bbb Q} a$ (since $p _{\Bbb Q} q <_{\Bbb Q} 0$), and $a \in r$. To make the arguments simpler, it is often easier to define "positive real numbers first", then add the cut 0, then define "formal negatives" (a negative real number is the COMPLEMENT of the closure (in $\Bbb Q$) of the set: $r = \{q: q \in r\}$ where $r$ is a positive real number (that is $r$ contains some positive rational number). We need to take the complement of the closure, instead of just the complement, to handle the case where $r$ contains its least upper bound). 

Tags 
addition, cuts, dedekind, proof 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proof of Addition Operation  jstarks4444  Number Theory  0  October 27th, 2011 03:23 PM 
Proof of addition formulae for sin(x + y) and cos(x + y)  maximus101  Real Analysis  1  March 18th, 2011 01:16 AM 
Sequence Addition Proof  jstarks4444  Number Theory  1  March 2nd, 2011 01:26 AM 
Need some help with a Dedekind domain proof  norway9k  Abstract Algebra  2  November 22nd, 2010 04:53 AM 
addition with parenthesis proof  ElMarsh  Algebra  7  September 8th, 2009 08:20 AM 