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Thread: Hausdorff space

  1. #1
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    Hausdorff space

    Show that X is a Hausdorff space iff. the diagonal $\displaystyle \Delta=\{x\times x:x\in X\}$ is a closed set in the product space.

    $\displaystyle \Rightarrow$
    Suppose X is a Hausdorff space. Then $\displaystyle \forall x_1,x_2\in X$ such that $\displaystyle x_1\neq x_2$ there exists neighborhoods $\displaystyle U_1 \ \text{and} \ U_2$ of $\displaystyle x_1, x_2$, respectively, such that $\displaystyle U_1\cap U_2=\O$.

    I don't know how I am supposed to use this to conclude the other piece.
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    Re: Hausdorff space

    Quote Originally Posted by dwsmith View Post
    Show that X is a Hausdorff space iff. the diagonal $\displaystyle \Delta=\{x\times x:x\in X\}$ is a closed set in the product space.
    If $\displaystyle (x,y)$ is such that $\displaystyle x\ne y$ can you show that there is a ball centered at $\displaystyle (x,y)$ that contains no point of $\displaystyle \Delta$.
    Hint: $\displaystyle \frac{|x-y|}{4}>0.$
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    Re: Hausdorff space

    Quote Originally Posted by Plato View Post
    If $\displaystyle (x,y)$ is such that $\displaystyle x\ne y$ can you show that there is a ball centered at $\displaystyle (x,y)$ that contains no point of $\displaystyle \Delta$.
    Hint: $\displaystyle \frac{|x-y|}{4}>0.$
    I don't think I can use that. I don't know it is a Metric space.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Hausdorff space

    Quote Originally Posted by dwsmith View Post
    Show that X is a Hausdorff space iff. the diagonal $\displaystyle \Delta=\{x\times x:x\in X\}$ is a closed set in the product space.

    $\displaystyle \Rightarrow$
    Suppose X is a Hausdorff space. Then $\displaystyle \forall x_1,x_2\in X$ such that $\displaystyle x_1\neq x_2$ there exists neighborhoods $\displaystyle U_1 \ \text{and} \ U_2$ of $\displaystyle x_1, x_2$, respectively, such that $\displaystyle U_1\cap U_2=\O$.

    I don't know how I am supposed to use this to conclude the other piece.
    Merely note that if $\displaystyle \Delta_X$ is closed, then $\displaystyle X^2-\Delta_X$ is open, and so if $\displaystyle x\ne y\in X$ then $\displaystyle (x,y)\in X^2-\DeltaX$ and so there exists some basic open neighbohood $\displaystyle U\times V$ of $\displaystyle (x,y)$ with $\displaystyle U\times V\subseteq X^2-\Delta_X$. Conclude from this that $\displaystyle x\in U$, $\displaystyle y\in V$, and $\displaystyle U\cap V=\varnothing$.

    Now, try to reverse this to get the reverse implication you asked about.
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