Hausdorff space

**dwsmith** · September 19th 2011, 02:59 PM

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

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

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

**Plato** · September 19th 2011, 03:22 PM

Originally Posted by dwsmith

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

If $(x,y)$ is such that $x\ne y$ can you show that there is a ball centered at $(x,y)$ that contains no point of $\Delta$ .
Hint: $\frac{|x-y|}{4}>0.$

**dwsmith** · September 19th 2011, 03:30 PM

Originally Posted by Plato

If $(x,y)$ is such that $x\ne y$ can you show that there is a ball centered at $(x,y)$ that contains no point of $\Delta$ .
Hint: $\frac{|x-y|}{4}>0.$

I don't think I can use that. I don't know it is a Metric space.

**Drexel28** · September 19th 2011, 04:16 PM

Originally Posted by dwsmith

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

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

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

Merely note that if $\Delta_X$ is closed, then $X^2-\Delta_X$ is open, and so if $x\ne y\in X$ then $(x,y)\in X^2-\DeltaX$ and so there exists some basic open neighbohood $U\times V$ of $(x,y)$ with $U\times V\subseteq X^2-\Delta_X$ . Conclude from this that $x\in U$ , $y\in V$ , and $U\cap V=\varnothing$ .

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

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