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Math Help - X_{\alpha} is Hausdorff

  1. #1
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    X_{\alpha} is Hausdorff

    If each X_{\alpha} is a Hausdorff space, then \prod X_{\alpha} is a Hausdorff space in both the box and product top.

    How do I go about proving this?

    I know since X_{\alpha} is Hausdorff, if for all x_1,x_2\in X_{\alpha} where x_1\neq x_2 there exists neighborhoods U_1 \ \text{and} \ U_2 of x_1 \ \text{and} \ x_2 that are disjoint.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: X_{\alpha} is Hausdorff

    Look here here. Ask your doubts.
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  3. #3
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    Opalg's Avatar
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    Re: X_{\alpha} is Hausdorff

    If (x_\alpha) and (y_\alpha) are distinct points in the product space, then they must differ in at least one coordinate. So there exists \alpha_0 say, such that x_{\alpha_0} \ne y_{\alpha_0}. Use the Hausdorff property in the \alpha_0-coordinate space to separate x_{\alpha_0} and y_{\alpha_0} by disjoint neighbourhoods. Then use those neighbourhoods to construct neighbourhoods in the product space that separate (x_\alpha) and (y_\alpha). The same construction will work for both the product and the box topologies.
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