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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- Mar 9th 2018, 10:15 AMhediWell 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. - Mar 9th 2018, 10:33 AMSlipEternalRe: 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. - Mar 9th 2018, 03:07 PMhediRe: 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 - Mar 10th 2018, 07:58 AMSlipEternalRe: Well ordered subset of r