Prove that an infinite chain contains a chain isomorphic to N or to -N

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SUMMARY

The discussion centers on proving that an infinite chain contains a chain isomorphic to either the natural numbers (N) or the negative integers (-N). The user proposes constructing a numerable chain (Y) from an infinite chain (X) by selecting elements based on their order. They outline two cases: one where elements greater than a chosen element x are selected, resulting in a strictly increasing sequence, and another where elements less than x are chosen, leading to a decreasing sequence. The user seeks assistance in formalizing their proof and justifying their steps to demonstrate the required bijection.

PREREQUISITES
  • Understanding of infinite chains in order theory
  • Familiarity with bijections and isomorphisms
  • Knowledge of natural numbers (N) and negative integers (-N)
  • Basic proof-writing skills in mathematics
NEXT STEPS
  • Study the properties of infinite chains in order theory
  • Learn about bijective functions and their role in establishing isomorphisms
  • Review formal proof techniques in mathematics
  • Explore examples of isomorphic structures in set theory
USEFUL FOR

Mathematics students, particularly those studying order theory, set theory, or proof techniques, will benefit from this discussion. It is also relevant for educators seeking to understand common student challenges in formal proof writing.

mahler1
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1. Homework Statement .
Prove that an infinite chain contains an a chain isomorphic (with the order) to N (the natural numbers) or to -N (negative integers).

3. The Attempt at a Solution .

I think I know how to solve the problem but I have problems to write a formal proof. I want to construct a numerable chain (call it Y) from the original one (call it X) in order to have a bijection that preserves the order.

I choose an arbitrary element x from X. As the set X is an infinite chain, then, then there must be infinite elements that are greater than x or x must be greater that infinite elements of X. If I'm in case 1, then I choose the first element x_1 such that x<x_1. I apply the same argument to x_1 and following the same step for for xn, I get a strictly increasing sequence of elements. I have the bijection I was looking for: {x_n} is stricly increasing, so x_i<x_j then i<j. If I'm in case 2, then there are infinite elements x' such that x'<x. I choose the first (first in terms of order) element x' such that x'<x. I call it x_1. I apply the same step for x_n in order to get a decreasing sequence {x_n}. If x_i<x_j, then i>j and -i<-j. The function from -N to {x_n} where f(i)=x_(-i) is a bijective function that preserves the order

Now, I don't know if my idea is correct or even comprehensible. In this case, I need help to write it formally, I feel that this is just the sketch of a proof but I have problems to justify all my steps and arguments.
 
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The argument of the proof looks fine to me. Indeed showing that there is a bijection between the sets you're constructing and the positive and negative natural numbers is the idea here (Isomorphisms must preserve).
 
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