I'm trying to prove the well-known identity
$$ \partial(A\times B)=(\partial A\times\overline B)\cup (\overline A\times \partial B), $$
where $A\subset X$ and $B\subset Y$. I have already shown the inclusion '$\supset$'. For the other inclusion, I have written the following:
Let $(x,y)\in \partial(A\times B)$ and assume to the contrary that $(x,y)\notin \partial A\times \overline B$ and $(x,y)\notin \overline A\times \partial B$. Since $(x,y)$ is a boundary point, $(U\times V)\cap (A\times B)\neq\emptyset$ for every open neighborhood $U\times V\ni (x,y)$. However, this means that $U\cap A\neq\emptyset$ for every open $U\ni x$. This implies that $x\in\overline A$, which is a contradiction, since $(x,y)\notin \overline A\times\partial B$.
The problem here is that I only used part of the antithesis, namely that $(x,y)\notin \overline A\times\partial B$. I could replicate the above "proof" to show that $\partial(A\times B)\subset \overline A\times\partial B$, which is simply not true. Something's amiss here, but I can't figure out what. All I can think of is that the antithesis is wrong, that there should be an 'or' instead of an 'and', but I can't see why this should be so.
Best Answer
The mistake is that $x \in \overline{A}$ does not contradict $(x,y) \notin \overline{A} \times \partial B$. However, from $x \in \overline{A}$ and $(x,y) \notin \overline{A} \times \partial B$ you can conclude that $y \notin \partial B$. Can you continue from there?