Is there an order-embedding from \mathbb Z^\infty to \mathbb Q for my paper?

  • Context: Graduate 
  • Thread starter Thread starter heptheorist
  • Start date Start date
Click For Summary
SUMMARY

This discussion centers on the existence of an explicit order-embedding from \(\mathbb{Z}^\infty\) to \(\mathbb{Q}\). The proposed function \(f: \mathbb{Z} \to (0,1) \cap \mathbb{Q}\) defines specific mappings for integers, allowing for the ordering of individual elements. The embedding function \(g: \mathbb{Z}^\infty \to \mathbb{Q}\) is constructed using a summation of the outputs from \(f\) and the differences \(d(n)\) between successive values. This method effectively demonstrates the desired order-preserving function, confirming the order isomorphism between the two sets.

PREREQUISITES
  • Understanding of order-embeddings and order-preserving functions
  • Familiarity with the structure of \(\mathbb{Z}^\infty\) and \(\mathbb{Q}\)
  • Knowledge of rational number properties and their representation
  • Basic concepts of infinite sums and convergence
NEXT STEPS
  • Research the properties of order-preserving functions in set theory
  • Explore the implications of order isomorphism between different mathematical structures
  • Study the construction and properties of infinite direct sums in algebra
  • Investigate further applications of rational mappings in mathematical analysis
USEFUL FOR

Mathematicians, particularly those focused on set theory, order theory, and algebra, will benefit from this discussion. It is also relevant for researchers exploring embeddings and isomorphisms in mathematical structures.

heptheorist
Messages
2
Reaction score
0
For a paper I'm writing: Does anyone know of an explicit order-embedding (i.e. an order-preserving function) from [itex]\mathbb Z^\infty[/itex], the direct sum of infinitely many copies of the integers ordered lexicographically, to [itex]\mathbb Q[/itex], the rationals? It need not be a surjective embedding, but that would be a plus (obviously the two sets are order isomorphic).
 
Physics news on Phys.org
What about this?

First, define a function [itex]f: \mathbb Z \to (0,1) \cap \mathbb Q[/itex], something like
0 -> 1/2
n -> 1 - 1/(2n) for i>0
n -> -1/(2n) for i<0

This allows to order individual "letters" (I like the analogy, I will keep it).
Use this function to evaluate the position of all 1-letter-words. In addition, the space to the next word can be used for all words beginning with this letter, in a similar way (0,1) was used for 1-letter-words. Let d(n)=f(n+1)-f(n) be this space.

Now, let [itex]g: \mathbb Z^\infty \to \mathbb Q[/itex] with
[tex]g(a_0,a_1,...) = \sum_i \left(f(a_i) \prod_{n=0}^{i-1} d(a_n)\right)[/tex]

I hope this works...
As the sum adds up a finite number of non-zero values, the result is rational.
 
This is a great idea, exactly what I wanted. Thank you!
 

Similar threads

  • · Replies 28 ·
Replies
28
Views
6K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 86 ·
3
Replies
86
Views
14K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 67 ·
3
Replies
67
Views
12K