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Thread: Well ordered subset of r

  1. #1
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    Well ordered subset of r

    Hey,
    Let A be a subset of R ,I is a well order on A,S is the ordinary order on A.
    Define the order J on A as the intersection of I and S.
    Prove that if A is full ordered by J then A is countable.
    Thank's in advance.
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  2. #2
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    Re: Well ordered subset of r

    There are many methods of proof for a proposition like this. Examples: 1. Direct proof. 2. Proof by contradiction. These two methods are very similar.

    In the first method, you would assume that A is full-ordered by J. Then use that to show that A is countable.
    With the second method, you would assume that A is full-ordered by J. Then, you would assume that A is uncountable. Then, you would try to reach a logical contradiction.

    If you have tried both of these methods and still gotten stuck, please post what you have tried so that we can see where to help you out.
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  3. #3
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    Re: Well ordered subset of r

    I used the property that every element in A has the first accessor in I that is also greater in S.So A is included in a countable union of countable set.I would like to see an elaborate proof in order to see that I don't miss something .
    Thank's in advance
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  4. #4
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    Re: Well ordered subset of r

    Quote Originally Posted by hedi View Post
    I used the property that every element in A has the first accessor in I that is also greater in S.So A is included in a countable union of countable set.I would like to see an elaborate proof in order to see that I don't miss something .
    Thank's in advance
    I can help verify a proof that you have written, but I am not going to provide a proof for you. If you have a proof to show, we can help make sure that you don't miss something.
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