# 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?

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May be you want $X$ to be Hausdorff? – Jyrki Lahtonen Jul 2 '11 at 8:37
@ Jyrki Lahtonen: yes $X$ has all the nice properties. – palio Jul 2 '11 at 8:58
You should be a bit more specific than "all the nice properties." In any case, try building an open neighborhood about a point not in $\Delta$ that is disjoint from $\Delta$. – user92843 Jul 2 '11 at 9:00
You might try proving first that the diagonal in $X^2$ is closed if $X$ is Hausdorff; it's basically the same argument, but with a little less clutter to confuse you. (It's actually true that $\Delta$ is closed iff $X$ is Hausdorff.) If $x \ne y$, use disjoint nbhds of $x$ and $y$ to build a nbhd of $(x,y)$ disjoint from $\Delta$ in $X^2$. Then generalize to $X^3$. – Brian M. Scott Jul 2 '11 at 9:11

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.

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wonderful answer!! thanks alot. – palio Jul 2 '11 at 9:25
Your last line (that the points on which two continuous functions agree forms a closed set) is really just a restatement of the fact that the diagonal in $X^2$ is closed. And once you know that, you can immediately conclude that each diagonal $\Delta_n \subseteq X^n$ is closed by the formula $\Delta_{n+1} = (\Delta_n \times X) \cap (X \times \Delta_n)$. – user92843 Jul 2 '11 at 10:07
@user92843: I agree, palio wanted continuous functions, but we don't need of that. – Davide Giraudo Jul 2 '11 at 13:52

It's not true in general. For example, it's not true if the topology on $X$ is indiscrete.

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Well, provided $X$ has at least two points. – user92843 Jul 2 '11 at 8:35