closed subspace of a cartesian product let $X$ be a topological space. let 
$$\Delta =\{(x,x,x)\, | x\in X\}\subset X^3$$
why $\Delta$ is closed in $X^3$? can we write a continuous map $f$ such that 
$\Delta $ is the preimage of some closed subset  or is there another argument?
 A: The result is true if and only if $X$ is Hausdorff (I guess we put the product topology on $X^3$). 
If $X$ is Hausdorff and $(x_1,x_2,x_3)\notin \Delta$ then we have $x_1\neq x_2$ or $x_2\neq x_3$ or $x_3\neq x_1$. We assume that $x_1\neq x_2$ (the other cases are similar). Let $U$ and $V$ two open disjoint subsets of $X$ such that $x_1\in U$ and $x_2\in V$. We have $U\times V\times X\subset \complement_{X^3}\Delta$ and $(x_1,x_2,x_3)\in U\times V\times X$. 
If $\Delta$ is closed, let $(x_1,x_2)\in X^2$ such that $x_1\neq x_2$. Since $(x_1,x_2,x_1)\notin\Delta$, we can find an open set $V$ in $X^3$ such that $(x_1,x_2,x_1)\in V$ and $V\subset \complement_{X^3}\Delta$. By definition of the product topology we can find two open subsets $V_1$ and $V_2$ such that $V_1\times V_2\times V_1\subset\complement_{X^3}\Delta$ and $x_1\in V_1$, $x_2\in V_2$, and we are done since if $x\in V_1\cap V_2$ then $(x,x,x)\in\complement_{X^3}\Delta$.
If you want to use continuous maps, let $p_i:X^3\rightarrow X$ the projections. 
We have $\Delta=\left\{(x_1,x_2,x_3)\in X^3,p_1(x_1,x_2,x_3)=p_2(x_1,x_2,x_3)=p_3(x_1,x_2,x_3)\right\}$ and you have to use the following result: if $Y_1$ and $Y_2$ are Hausdorff spaces and $f,g:Y_1\rightarrow Y_2$ are continuous then the set $\left\{x\in Y_1,f(x)=g(x)\right\}$ is closed.
A: It's not true in general.  For example, it's not true if the topology on $X$ is indiscrete.
