Proving countable ordinal embeds in R

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Homework Statement


show that if q is any countable ordinal, then there is a countable set A ⊆ R (in fact we can require A ⊆ Q), so that (A, <) ∼= (q, ∈).

The Attempt at a Solution


since q is a countable ordinal this implies that it has a mapping to the naturals.
to me this seems strong enough. and its also well ordered.
 
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It's hard to tell from the notation you've used, but I suspect the question is asking you to prove there is an order-preserving mapping.
From countability of q you can get a bijection to the naturals, but in most cases it won't be order preserving.
 
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in order to preserve order, can we use a permutation group to create a linear order
 
The permutation groups I am familiar with are on finite sets. Unless q is finite - in which case the entire problem becomes trivial - I can't see how a permutation on a finite number of elements would help. I think you will need to use the structure of countable ordinals to prove this.
 
Use transfinite induction and the fact that ##\mathbb{R}## is order isomorphic to any interval ##(a,b)##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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