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Let $X$ be a topological space. The diagonal of $X times X$ is the subset $$D = {(x,x)in Xtimes Xmid x in X}.$$

Show that $X$ is Hausdorff if and only if $D$ is closed in $X times X$.

First, I tried to show that $X times X setminus D$ is open using the fact that $X times X$ is Hausdorff (because $X$ is Hausdorff), but I couldn't find an open set that contains a point outside $D$ and is disjoint to it...

You have two things to show: that if $D$ is closed, then $X$ is Hausdorff, and that if $X$ is Hausdorff, then $D$ is closed.

Suppose first that $D$ is closed in $Xtimes X$. To show that $X$ is Hausdorff, you must show that if $x$ and $y$ are any two points of $X$, then there are open sets $U$ and $V$ in $X$ such that $xin U$, $yin V$, and $Ucap V=varnothing$. The trick is to look at the point $p=langle x,yranglein Xtimes X$. Because $xne y$, $pnotin D$. This means that $p$ is in the open set $(Xtimes X)setminus D$. Thus, there must be a basic open set $B$ in the product topology such that $pin Bsubseteq(Xtimes X)setminus D$. Basic open sets in the product topology are open boxes, i.e., sets of the form $Utimes V$, where $U$ and $V$ are open in $X$, so let $B=Utimes V$ for such $U,Vsubseteq X$. Can you see how to use this to get the desired open neighborhoods of $x$ and $y$?

Now suppose that $X$ is Hausdorff. To show that $D$ is closed in $Xtimes X$, you need only show that $(Xtimes X)setminus D$ is open. To do this, just take any point $pin(Xtimes X)setminus D$ and show that it has an open neighborhood disjoint from $D$. I suggest that you try to reverse what I did above. First, $p=langle x,yrangle$ for some $x,yin X$, and since $pnotin D$, $xne y$. Now use disjoint open neighborhoods of $x$ and $y$ to build a basic open box around $p$ that is disjoint from $D$.

There is a related problem:

Let $f,g:Ato B$ be two continuous maps, $B$ is a Hausdorff space and $A$ is an arbitrary space. Then the set $C(f,g)={xin A| f(x)=g(x)}$ is the coincidence set of $f$ and $g$. Prove $C(f,g)$ is closed.

to proof this, for every point $x_0$ not in $C(f,g)$, $y_f=f(x_0)$ and $y_g=g(x_0)$ are two distinguished point in $B$, because $B$ is Hausdorff, you get two disjoint open neighborhood $U(y_f)$ and $U(y_g)$, because $f,g$ is continuous, $O=f^{-1}(U(y_f))cap g^{-1}(U(y_g))$ is open in $A$. You can verify that $O$ is a neighborhood of $x_0$ disjoint of $C(f,g)$, for if $cin C(f,g)$ and $cin O$, then $cin f^{-1}(U(y_f))$ and $cin g^{-1}(U(y_g))$, this is equal to say $f(c)=g(c)$ is in both $U(y_f)$ and $U(y_g)$, which is impossible because the two neighborhood is disjoint.

Set $A=Xtimes X$, $B=X$, $f=pi_1$, $g=pi_2$ be the projection map, you get the desired result.

Hope this can illustrate the problem you ask.

Let X be Hausdorff, then if $xne y$ there are neighborhoods $V_x$ and $V_y$ such that $V_x cap V_y = emptyset$. Therefore $V_xtimes V_y cap D=emptyset$ and the complement of $D$ is open. Now, assume that the latter is true. Then, for any point $(x,y)$, $xne y$, there is an open set around it that does not intersect $D$. Therefore, there are two sets $xin V_x$ and $yin V_y$ such that $V_xtimes V_y$ doesn't intersect $D$, therefore $V_x cap V_y = emptyset$.

Let $(x,y)in Xtimes X$, with $xneq y$. Since $x,yin X$, $xneq y$, and $X$ is Hausdorff, there exist open set $mathcal{U}$ and $mathcal{V}$ such that $xinmathcal{U}$, $yinmathcal{V}$, and $mathcal{U}capmathcal{V}=varnothing$. Now, $mathcal{U}timesmathcal{V}$ is an open subset of $Xtimes X$. It contains $(x,y)$. Does it intersect $D$?

Conversely, suppose $D$ is closed, and let $x,yin X$, $xneq y$. Then $(x,y)notin D$, so there is an open subset $mathscr{O}$ such that $(x,y)inmathscr{O}subseteq Xtimes X - D$. Do you know something about open sets of a special type in $Xtimes X$ that might let you obtain open sets of $X$ $mathcal{U}$ and $mathcal{V}$ as in the previous paragraph?