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Math Help - Simply ordered set

  1. #1
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    Simply ordered set

    Every simply ordered set is a Hausdorff space in the order topology.

    Since it is simply ordered, the relation is reflexive, anti-symmetric, and transitive. How can this coupled with the Hausdorff condition help show that it is the order topology?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Simply ordered set

    Quote Originally Posted by dwsmith View Post
    Every simply ordered set is a Hausdorff space in the order topology.

    Since it is simply ordered, the relation is reflexive, anti-symmetric, and transitive. How can this coupled with the Hausdorff condition help show that it is the order topology?
    You are misinterpreting the question. It's saying that if X is a space where you haven given it the order topology for some total ordering then the resulting topological space is Hausdorff.
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    Re: Simply ordered set

    Quote Originally Posted by Drexel28 View Post
    You are misinterpreting the question. It's saying that if X is a space where you haven given it the order topology for some total ordering then the resulting topological space is Hausdorff.
    Ok. However, I am not sure how to do that one either.
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    MHF Contributor Drexel28's Avatar
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    Re: Simply ordered set

    Quote Originally Posted by dwsmith View Post
    Ok. However, I am not sure how to do that one either.
    Let x,y\in X be distinct. Since < is a total ordering on X we may assume without loss of generality that x<y. We have two choices, either there exists z with x<z<y in which case take U=(-\infty,z) and V=(z,\infty), else (x,y) is empty and take U=(-\infty,y) and V=(x,\infty). Regardless, U and V are disjoint neighborhoods of x,y respectively. The conclusion follows.
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