# Thread: Well ordered subset of r

1. ## 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.

2. ## 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.

3. ## 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 .