Determine interior and boundary of $A\times B$ Let $(X,\mathcal{O}_X)$ and $(Y,\mathcal{O}_Y)$ be topological spaces and $A,B$ subset of $X,Y$ respectively. I have to find the interior $(A\times B)^\circ$ and the boundary $\partial(A\times B)$. Furthermore, it is that the open sets are given by the product topology on $A\times B$.
This sounds pretty easy, nevertheless I am not quite sure what exactly I should show. The interior and boundary both depend on the topology itself. Is this context to general to actually "show" something? 
 A: (To remove the question from unanswered).
According to Exercise 2.3.B from [Eng], $(A\times B)^\circ=A^\circ\times B^\circ$ and $\partial(A\times B)=\partial A\times \overline{B}\cup \overline{A}\times \partial B$. Prove this.
By Proposition 2.3.1 from [Eng], the set $A^\circ\times B^\circ$ is open, so $A^\circ\times B^\circ\subset (A\times B)^\circ$. On the other hand, let $(x,y)\in (A\times B)^\circ$ be any point. Then there exists an element $U\times V$ of the canonical base at $X\times Y$ such that $(x,y)\in U\times V\subset  A\times B$. Then $U\subset A$ and $V\subset B$. Since $U$ and $V$ are open in $X$ and $Y$, respectively, we have $U\subset A^\circ$ and  $V\subset B^\circ$. Then $(x,y)\in A^\circ\times B^\circ$. 
By Proposition 2.3.3 from [Eng], $\overline{A\times B}=\overline A\times \overline B$, so 
$$\partial(A\times B)=$$ $$\overline{A\times B}\setminus (A\times B)^\circ=$$
$$\overline A\times \overline B \setminus A^\circ\times B^\circ =$$ $$
((\partial A\cup A^\circ)\times (\partial B\cup B^\circ)) \setminus A^\circ\times B^\circ=$$ $$\partial A\times \partial B\cup \partial A\times B^\circ \cup A^\circ \times \partial B=$$ $$(\partial A\times \partial B\cup \partial A\times B^\circ)\cup 
(\partial A\times \partial B\cup A^\circ \times \partial B)=$$ $$ \partial A\times \overline{B}\cup \overline{A}\times \partial B.$$
References
[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.





