MHB Every interval (a,b) contains both rational and irrational numbers

Click For Summary
The discussion revolves around understanding Theorem 7 from Charles Chapman Pugh's "Real Mathematical Analysis," specifically how the condition \( a < b \) implies the existence of two distinct rational numbers in the set \( B \setminus A \). Participants seek clarification on the notation and concepts related to Dedekind cuts, particularly the notation \( A|A' \). The conversation emphasizes the definition of a Dedekind cut as a pair of subsets that satisfy specific conditions. There is a focus on ensuring clarity regarding the implications of the theorem and the associated notation. Overall, the thread aims to deepen understanding of the relationship between rational and irrational numbers within the context of real analysis.
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Chapter 1:"Real Numbers" of Charles Chapman Pugh's book "Real Mathematical Analysis.

I need help with the proof of Theorem 7 on pages 19-20.

Theorem 7 (Chapter 1) reads as follows:
View attachment 3828
View attachment 3829In the above proof, Pugh writes:

" ... ... The fact that $$a \lt b$$ implies the set B \ A contains two distinct rational numbers, say $$r, s$$. ... ... "

Can someone help me to understand exactly how it follows that $$a \lt b$$ implies the set B \ A contains two distinct rational numbers, say $$r, s$$?

Peter***NOTE***

Since Theorem 7, Chapter 1, mentions cuts, i am providing Pugh's definition of a Dedekind cut, as follows:View attachment 3830
 
Last edited:
Physics news on Phys.org
Peter, could you please explain the meaning of the notation $A|A'$?
 
Euge said:
Peter, could you please explain the meaning of the notation $A|A'$?
Sorry Euge, I should have included that notation after the definition of a cut in $$\mathbb{Q}$$ ... ... my apologies ...

A cut in $$\mathbb{Q}$$ is a pair of subsets $$A, B$$ with the three conditions shown above in my post ... the Dedekind cut is denoted $$A|B$$ ...

So $$A|A$$' is a Dedekind cut involving the two sets $$A$$ and $$A'$$

Hope that helps ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
6
Views
5K
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K