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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.