# Show that the diagonal $\Delta$ is closed in $X\times X$

I'm trying to show the following implication:

Given $(X,\mathcal{T})$

$\bigcap_{U \in \mathcal{U}} \bar{U} = \{x\} \quad \forall x \in X; \text{ where }\mathcal{U}\text{ is the set of all open neighbourhoods of }x \rightarrow$ The diagonal $\Delta = \{(x,x)|x\in X\}$is closed in $X \times X$.

I tried to show that all points of closure of $\Delta$ lie in $\Delta$, but I could only get as far as this:

If $(x,y) \in \bar{\Delta}, \text{ then }\bar{U(x)} \cap \bar{V(y)} \neq \phi \quad \forall \{U(x),V(y)\} \subset \mathcal{T}$.

But I can't think of a way to show that implies $x=y$ without using the fact that the space is Hausdorff, which I'm trying to avoid.

I was trying to think of a way to show that $\bigcap_{\{U(x),V(y)\} \subset \mathcal{T}}[U(x)\cap V(y)]$ is nonempty, since that implies that $\{x\}=\{y\}$, but can't see an argument to show that would be so.

Let $x$ and $y$ be two distinct points of $X$. According to the hypotheses of the problem, there is an open neighborhood $U$ of $x$ such that $y\not\in \overline{U}$. Therefore if $V=X\setminus\overline{U}$ then $U$ is an open neighborhood of $x$, $V$ is an open neighborhood of $y$, and $U\cap V=\emptyset$.
The set $U\times V$ is an open neighborhood of $(x,y)$ in $X\times X$, and cannot contain any point of $\Delta$ because $U$ and $V$ are disjoint. Therefore $U\times V\subset (X\times X)\setminus\Delta$.
As $(x,y)$ was any point of $(X\times X)\setminus\Delta$, this shows that $(X\times X)\setminus\Delta$ is open, so $\Delta$ is closed.
• Note that the hypothesis implies that $X$ is Hausdorff, as seen in carmichael561's answer. The converse also holds : $X$ is Hausdorff iff $\Delta$ is closed in $X^2.$ Commented Mar 1, 2016 at 7:57