Proof that $\partial(A\times B) = ((\partial A)\times B)\cup (A\times(\partial B))$

I need to prove that:

$$\partial(A\times B) = ((\partial A)\times B)\cup (A\times(\partial B))$$

In other terms, the boundary of the cartesian product is the union of the things in the RHS.

I've found this question but it does not even provide an intuition.

If $x\in ((\partial A)\times B)\cup (A\times(\partial B))$, then certainly $x \in \partial(A\times B)$ because:

$$((\partial A)\times B)\cup (A\times(\partial B))$$

is the union of $$(\partial A, something)$$

with

$$(something, \partial B)$$ Any ideads on how to say it rigorously?

What about the other side of the proof? What does this result means?

Solution 1:

This is not the right conclusion since LHS is a closed set and RHS may be not closed. We prove the right result: $$ \partial(A\times B) = (\partial A\times \overline{B})\cup (\overline{A}\times\partial B) $$ Proof: \begin{align} \partial(A\times B) &=\overline{A\times B}-(A\times B)^o \\ &=\overline{A}\times\overline{B}-A^o\times B^o\tag1 \\ &=\overline{A}\times\overline{B}\:\cap (A^o\times B^o)^c \\ &=\overline{A}\times\overline{B}\:\cap ((A^o)^c\times X)\cup(X\times (B^o)^c)\tag2 \\ &=((\overline{A}\times\overline{B})\:\cap ((A^o)^c\times X))\cup((\overline{A}\times\overline{B})\:\cap ((X\times (B^o)^c)) \\ &=((\overline{A}\cap (A^o)^c)\times(\overline{B}\cap X))\cup((\overline{A}\cap X)\times(\overline{B}\cap (B^o)^c))\tag3 \\ &=((\overline{A}-A^o)\times\overline{B})\cup(\overline{A}\times(\overline{B}- B^o)) \\ &=(\partial A\times \overline{B})\cup(\overline{A}\times \partial B) \end{align} $(1)$ is by the following formulas $$ \overline{A\times B}=\overline{A}\times\overline{B}\quad\text{and }\quad (A\times B)^o=A^o\times B^o $$ $(2)$ is by the following formula $$ (A\times B)^c=(A^c\times X)\cup(X\times B^c) $$ $(3)$ is by the following formula $$ (A\times B)\cap(C\times D)=(A\cap C)\times (B\cap D) $$