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

In summary, the problem is to prove that an infinite chain contains a chain isomorphic to either the set of natural numbers or the set of negative integers. The attempt at a solution involves constructing a countable chain from the original one, using a bijection that preserves the order. The author has provided a sketch of the proof, but is unsure of how to write it formally and is seeking assistance.
  • #1
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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  • #2
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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Related to Prove that an infinite chain contains a chain isomorphic to N or to -N

1. What does it mean for a chain to be isomorphic to N or -N?

Isomorphism is a mathematical concept that describes a one-to-one correspondence between two mathematical structures. In this context, a chain isomorphic to N or -N means that there exists a bijection between the elements of the infinite chain and the natural numbers (N) or the negative integers (-N).

2. Why is it important to prove that an infinite chain contains a chain isomorphic to N or -N?

This proof is important because it helps us understand the structure of infinite chains and how they relate to other mathematical concepts. It also has applications in various branches of mathematics, such as group theory and topology.

3. How do you prove that an infinite chain contains a chain isomorphic to N or -N?

To prove this statement, we can use the well-ordering principle, which states that every non-empty set of natural numbers has a least element. By assuming that an infinite chain does not contain a chain isomorphic to N or -N, we can use this principle to construct a contradiction, thus proving the statement.

4. Can you provide an example of an infinite chain containing a chain isomorphic to N or -N?

Yes, a simple example would be the set of all integers (Z) with the natural ordering. In this case, we can define a bijection between Z and -N by mapping each integer to its negative counterpart, thus showing that Z contains a chain isomorphic to -N.

5. What are the implications of this proof in other areas of mathematics?

This proof has implications in various branches of mathematics, such as abstract algebra, topology, and number theory. It helps us understand the structure of infinite chains and how they relate to other mathematical structures. It also has applications in the study of infinite groups and topological spaces.

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