# Hausdorff space

• September 19th 2011, 02:59 PM
dwsmith
Hausdorff space
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.
• September 19th 2011, 03:22 PM
Plato
Re: Hausdorff space
Quote:

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: $\frac{|x-y|}{4}>0.$
• September 19th 2011, 03:30 PM
dwsmith
Re: Hausdorff space
Quote:

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.
• September 19th 2011, 04:16 PM
Drexel28
Re: Hausdorff space
Quote:

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.