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Math Help - A subspace of a Hausdorff space is Hausdorff

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    A subspace of a Hausdorff space is Hausdorff

    A subspace of a Hausdorff space is Hausdorff.

    Let X be Hausdorff and Y\subset X

    Since Y is a subspace, Y=X\cap C where C is some open set in X.

    Let x_1,x_2\in X\cap C

    X\cap C is the intersection of open sets so it is open. Therefore, x_1\in U \ \text{and} \ x_2\in V where U and V are open sets in X\cap C. Again since X\cap C is open, U and V can be constructed in such a manner that U\cap V=\O.

    Thus, the subspace is Hausdorff.

    Correct?
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    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    A subspace of a Hausdorff space is Hausdorff.

    Let X be Hausdorff and Y\subset X

    Since Y is a subspace, Y=X\cap C where C is some open set in X.

    Let x_1,x_2\in X\cap C

    X\cap C is the intersection of open sets so it is open. Therefore, x_1\in U \ \text{and} \ x_2\in V where U and V are open sets in X\cap C. Again since X\cap C is open, U and V can be constructed in such a manner that U\cap V=\O.

    Thus, the subspace is Hausdorff.

    Correct?
    This isn't quite correct. The fact that Y is a subspace of X says that given any open subset U\subseteq Y one has that U={\color{red}{Y}}\cap V for some open set [etx]V[/tex] in X. want to take this and try again?
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    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    A subspace of a Hausdorff space is Hausdorff.
    Let X be Hausdorff and Y\subset X
    Since Y is a subspace, Y=X\cap C where C is some open set in X.
    Suppose that a\in Y~,~b\in Y~\&~a\ne b. Then \{a,b\}\subset X.
    But that means there are two disjoint open sets in X such that a\in\mathcal{O}_a~\&~b\in\mathcal{O}_b WHY?
    What can you say about Y\cap\mathcal{O}_a~\&~Y\cap\mathcal{O}_b~?
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    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by Drexel28 View Post
    This isn't quite correct. The fact that Y is a subspace of X says that given any open subset U\subseteq Y one has that U={\color{red}{Y}}\cap V for some open set [etx]V[/tex] in X. want to take this and try again?
    From that, we know U is open in X since it is the intersection of two open sets in X. Let x_1,x_2\in U. Since U is open, disjoint sets V_1,V_2 can be constructed s.t. x_1\in V_1 \ \text{and} \ x_2\in V_2 and V_1\cap V_2=\O

    Does this do it then?
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    MHF Contributor Drexel28's Avatar
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    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    From that, we know U is open in X since it is the intersection of two open sets in X. Let x_1,x_2\in U. Since U is open, disjoint sets V_1,V_2 can be constructed s.t. x_1\in V_1 \ \text{and} \ x_2\in V_2 and V_1\cap V_2=\O

    Does this do it then?
    Once again, not quite. You need to start with two arbitrary points x_1,x_2\in Y and construct two disjoint neighborhoods of them. You know you can do this in X, in other words there exists disjiont open sets V_1,V_2 in X with x_i\in V_i, now how can you project these down to open subsets in Y which are disjoint and contain x_i?
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    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by Drexel28 View Post
    Once again, not quite. You need to start with two arbitrary points x_1,x_2\in Y and construct two disjoint neighborhoods of them. You know you can do this in X, in other words there exists disjiont open sets V_1,V_2 in X with x_i\in V_i, now how can you project these down to open subsets in Y which are disjoint and contain x_i?
    Let x_1,x_2\in Y; V_1,V_2\in X, and V_1\cap V_2=\O where V_1,V_2 are open in X.
    x_1\in V_1 \ \text{and} \ x_2\in V_2


    Y\cap V_1=U_1 \ \text{and} \ Y\cap V_2=U_2.

    U_1\cap U_2=\O

    Correct?
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    MHF Contributor Drexel28's Avatar
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    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    Let x_1,x_2\in Y; V_1,V_2\in X, and V_1\cap V_2=\O where V_1,V_2 are open in X.
    x_1\in V_1 \ \text{and} \ x_2\in V_2


    Y\cap V_1=U_1 \ \text{and} \ Y\cap V_2=U_2.
    Yes.
    U_1\cap U_2=\O

    Correct?
    Yes.
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