Diagonal is closed in $(X, \tau_1) \times (X, \tau_2)$ (both are Hausdorff)

Question

From Willard's General Topology:

Let $\tau_1$ and $\tau_2$ be Hausdorff topologies on the same set $X$. Let $\tau_3 = \tau_1 \cap \tau_2$. If $(X, \tau_3)$ is Hausdorff, then the diagonal is closed in $(X, \tau_1) \times (X, \tau_2)$.

I know that since $\tau_1, \tau_2,$ and $\tau_3$ are all Hausdorff, that the diagonal $\Delta = \{(x,x) : x \in X\}$ is closed in each of the following product topologies $(X, \tau_1) \times (X, \tau_1)$, $(X, \tau_2) \times (X, \tau_2)$, and $(X, \tau_3) \times (X, \tau_3)$.

Without thinking much about the question, it seems intuitive that yes, $\Delta$ will also be closed in $(X, \tau_1) \times (X, \tau_2)$. I've tried to proceed by contradiction below, but I can't figure out how to finish it off. I would appreciate any help finishing it off.

I tried to proceed by contradiction:

Supposing that $\Delta$ is not closed in $(X, \tau_1) \times (X, \tau_2)$, then $\exists y \in X, y \neq x$, such that $(x,y) \in \overline{\Delta}$ (the closure of $\Delta$ in $\tau_1 \times \tau_2$). Then $\forall U \times V$ basic nbhoods of $(x,y)$, we must have that $(U\times V) \cap \Delta \neq \emptyset$. So $\exists z \in X$, such that $(z,z) \in U \times V$.

Also, since $U \times V$ is a basic nbhood of $(x,y)$ in $(X, \tau_1) \times (X, \tau_2)$, then $U$ is a basic nbhood of $x$ in $(X, \tau_1)$ and $V$ is a basic nbhood of $y$ in $(X, \tau_2)$. But then, since $(z,z) \in U \times V$, we must have $z \in U$ and $z \in V$. So $U, V$ are not disjoint. But $U$ was an arbitrary basic nbhood of $x$ in $(X, \tau_1)$ and $V$ was an arbitrary basic nbhood of $y$ in $(X, \tau_2)$, so there cannot exist any open disjoint sets $A \in \tau_1$, $B \in \tau_2$ such that $x \in A$ and $y \in B$...

I'm guessing this contradicts the Hausdorff-ness of $(X, \tau_3)$, but I'm having trouble putting it together.

Thanks.

Answer 1

You are making your life overly complicated.

Take a point $(x,y)$ off the diagonal. We know there is an open set $U$ of $(X,\tau_3)$ and an open set $V$ of $(X,\tau_3)$ such that $x\in U$, $y\in V$, and $U\times V$ does not intersect the diagonal; i.e., $U\cap V=\varnothing$. This because $(X,\tau_3)$ is Hausdorff, so the diagonal in $X\times X$, with the $\tau_3\times\tau_3$ topology, is closed.

But $\tau_3=\tau_1\cap\tau_2$. Thus, $U\in \tau_1$ and $V\in \tau_2$. That means that $U\times V$ is open in $\tau_1\times\tau_2$, contains $(x,y)$, and does not intersect the diagonal.