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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.

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You need to give a few definitions: $\Delta$ is a subset of $X\times Y$? (not an element). And $U(x)$ (resp. $V(y)$) is an open neighbourhood of $x$ (resp. $Y$)? The Latex command for cartesian product is \times. – Michalis Feb 21 '13 at 16:04
Yes, sorry. It is a subset. Its infact the diagonal i.e $\Delta = \{(x,x) \mid x \in X\}$. – Fatsho Feb 21 '13 at 16:19
It seems your question is not very difficult. You just have to write it down properly, so people can help you. I would say @Michalis knows the answer and I probably do to. I suggest you improve your question... – André Caldas Feb 21 '13 at 16:20

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 $W\cap\Delta\ne\varnothing$.

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