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?
 A: 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
$$
(A\times B)\cap(C\times D)=(A\cap C)\times (B\cap D)
$$
