Q* (the set of rational cuts) has least upper bound property or not?

  • #1

Main Question or Discussion Point

I am struggling to draw this point home:

To prove that R has LUB property, we used the following reasoning:
First we defined R to be set of cuts (having certain properties) where each cut corresponds to a real number and then we proved any subset A of R has LUB (least upper bound) property.

If I use similar approach to define Q* to be the set of all rational cuts.
Then let A be the subset of Q*. Let Υ be the union of all cuts in A. Then this Υ is a cut in Q* and is supremum of A.

(Note, to define Q* above, we didn't need to introduce R, as we are playing with only rational cuts so far.)

Is above reasoning false? and if false, how to prove it.

If the reasoning is true, then the coclusion that Q* has LUB leads contrary to what we have read in the books (i.e. Q does not have LUB). Or the conclusion with the help of cuts that R has LUB has problem?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,794
925
Your reasoning is incorrect. In the set of all Dedekind cuts (defining the real numbers) we can show that the union of all cuts less a given cut is itself a Dedekind cut and so defines the least upper bound. If we take, for example, the set of all rational cuts x, such that [itex]x^2< 2[/itex], the union is not a rational cut.
 
  • #3
Thanks a lot! The sqrt 2 example above, makes the picture clear.
 

Related Threads for: Q* (the set of rational cuts) has least upper bound property or not?

Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
2K
Replies
5
Views
2K
Top