# 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?

• For an imtuition, consider the product of a closed disc and a closed interval. This is a cylinder. The boundary of the cylinder has three components. The top and bottom of the cylinder are the product of the disc with the endpoints of the interval. The curved main tube of the cylinder is the product of the whole interval with the circle that bounds the closed disc.
– MJD
Apr 10, 2016 at 15:10
• This is wrong for LHS is closed set and RHS may be not closed Apr 11, 2016 at 7:15

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 $$(A\times B)\cap(C\times D)=(A\cap C)\times (B\cap D)$$
• what is $A^c$?? Apr 11, 2016 at 15:55
• It is complement of $A$, ie $A^c=X-A$. Apr 11, 2016 at 19:41