Closure in cartesian product space

Question

If $a = (x,y)$ is a point of closure of $\Delta \subset X \times Y$, is it right to say that $U(x) \times V(y) \cap \Delta \neq \emptyset$.

I've studied closure only in single(?) spaces, so I essentially want to know if I can rewrite the definition as above for product spaces.

Answer 1

I think that you have the right idea, but what you want to say is this:

$\langle x,y\rangle\in\operatorname{cl}\Delta$ if and only if $(U\times V)\cap\Delta\ne\varnothing$ whenever $U$ is an open nbhd of $x$ in $X$ and $V$ is an open nbhd of $y$ in $Y$.

This is true because $\{U\times V:U\in\tau(X)\text{ and }V\in\tau(Y)\}$ is a base for the product topology on $X\times Y$, where $\tau(X)$ is the topology on $X$, and $\tau(Y)$ is the topology on $Y$.

That is, by definition $\langle x,y\rangle\in\operatorname{cl}\Delta$ if and only if $W\cap\Delta\ne\varnothing$ for every open nbhd $W$ of $\langle x,y\rangle$ in $X\times Y$. But if $W$ is an open nbhd $W$ of $\langle x,y\rangle$ in $X\times Y$, then by definition of the product topology there are $U\in\tau(X)$ and $V\in\tau(Y)$ such that $\langle x,y\rangle\in U\times V\subseteq W$, so if $(U\times V)\cap\Delta\ne\varnothing$, then certainly $W\cap\Delta\ne\varnothing$.