Importance of diagonal (topology) On p.101 of Munkres Topology, Excersise 13 states that a space $X$ is Hausdorff if and only if the diagonal $\Delta=\{(x,x):x\in X\}$ is closed on $X\times X$. 
To me, it seems a nice characterization of Hausdorff spaces. However, it doesn't seem to be the most useful tool to prove that a concrete space $X$ is Hausdorff. 
So my questions are the following: Is this characterization really useful on working with Hausdorff spaces? Is the diagonal set useful in some particular branch of mathematics? Thank you in advance.
 A: In algebraic geometry, the standard spaces are not Hausdorff in the sense that you can separate any two points, but they are Hausdorff using the closed diagonal condition. The most simple example is the space $\mathbb{C}$ with the Zariski topology - namely the closed set are just the zeros of some nonzero polynomial. Since every polynomial have only finitely many roots, the closed set are just the finite sets (and the whole space), so that the open sets are the cofinite sets. Clearly, the intersection of any two (non empty) open sets is non empty (or equivalently, the union of two closed set is finite and therefore cannot be the entire space).
On the other hand, the product space is just $\mathbb{C}^2$ and the diagonal is exactly the zeros of the polynomial $f(x,y)=x-y$.
A: Necessity: Let $(X,\tau)$ be a Hausdorff space and $(x,y)\notin\Delta$.
$$\left.\begin{array}{rr} (x,y)\notin\Delta\Rightarrow x\neq y  \\ \\ (X,\tau) \text{ is Hausdorff } \end{array}\right\}\Rightarrow (\exists U\in\mathcal{U}(x))(\exists\in\mathcal{U}(y))(U\cap V=\emptyset)$$
$$\overset{?}{\Rightarrow}$$
$$(U\times V\in\mathcal{U}(x,y))((U\times V)\cap \Delta=\emptyset)$$
$$\Rightarrow$$
$$(U\times V\in\mathcal{U}(x,y))(U\times V\subseteq \setminus \Delta)$$
$$\Rightarrow$$
$$(x,y)\in (\setminus \Delta)^{\circ}$$
Therefore $$\setminus\Delta\subseteq (\setminus\Delta)^{\circ}\ldots (1)$$ On the other hand 
$$(\setminus\Delta)^{\circ}\subseteq\setminus\Delta \ldots (2)$$ is always true. Then we have
$$(1),(2)\Rightarrow (\setminus\Delta)^{\circ}=\setminus\Delta \Rightarrow\setminus\Delta\in\tau\star\tau \Rightarrow\Delta\in \mathcal{C}(X\times X).$$
$------------------------------------$
Sufficiency: Let $\Delta\in \mathcal{C}(X\times X)$ and  let $x,y\in X$ and $x\neq y.$
$$\left.\begin{array}{rr} (x,y\in X)(x\neq y)\Rightarrow (x,y)\notin\Delta  \\ \\ \Delta \in \mathcal{C}(X\times X)\Rightarrow \overline{\Delta}=\Delta \end{array}\right\}\Rightarrow (\exists W\in\mathcal{U}(x,y))(W\cap \Delta=\emptyset)$$
$$\Rightarrow$$
$$(\exists \mathcal{A}_1\subseteq\tau_1)(\exists \mathcal{A}_2\subseteq\tau_2)((x,y)\in W=\cup_{(A_1\in\mathcal{A}_1)(A_2\in\mathcal{A}_2)}(A_1\times A_2))(W\cap \Delta=\emptyset)$$ 
$$\Rightarrow$$
$$(\exists U\in\mathcal{A}_1\subseteq\tau_1)(\exists V\in \mathcal{A}_2\subseteq\tau_2)(x\in U)(y\in V)((x,y)\in U\times V\subseteq W)(W\cap \Delta=\emptyset)$$
$$\Rightarrow$$
$$(U\in\mathcal{U}(x))(V\in \mathcal{U}(y))((U\times V)\cap \Delta=\emptyset)$$
$$\Rightarrow$$
$$(U\in\mathcal{U}(x))(V\in \mathcal{U}(y))(U\cap V=\emptyset).$$
