Let $A \subset \mathbb{R}^n, B \subset \mathbb{R}^m.$
I want to prove $\mathrm{int}(A\times B)=(\mathrm{int}A)\times (\mathrm{int} B)$.
I could prove $\mathrm{int}(A\times B)\supset (\mathrm{int}A)\times (\mathrm{int} B)$, but I'm difficulty in proving the reverse.
Let $x\in \mathrm{int}(A\times B)$.
Due to the definition of int$(A\times B),$ there exists an open set $W$ s.t. $x\in W\subset A\times B.$
Since $x\in W\subset A\times B,$ I can write $W=U\times V$, where $U\subset A, V\subset B$ and $x=(u,v)$ where $u\in U, v\in V.$
So, all that is left is to show $U,V$ are open.
Here are two questions of mine.
(i) How can I show $U,V$ are open ?
(ii) Is $ \big[ W\subset A\times B \Rightarrow W$ is written as $W=U\times V\big]$ in my proof correct ?
$\operatorname{Int}(A) \times \operatorname{Int}(B)$ is open and a subset of $A \times B$. As the interior is the largest open subset of a set, $\operatorname{Int}(A) \times \operatorname{Int}(B) \subseteq \operatorname{Int}(A \times B)$ is immediate. Also, as projections are open, $\pi_X[\operatorname{Int}(A \times B)]$ is an open subset of $A$ and by the same maximality $\pi_X[\operatorname{Int}(A \times B)] \subseteq \operatorname{Int}(A)$ and likewise $\pi_Y[\operatorname{Int}(A \times B)]$ is an open subset of $B$ from which $\pi_Y[\operatorname{Int}(A \times B)] \subseteq \operatorname{Int}(B)$ follows. Simple set theory then tell us $\operatorname{Int}(A \times B) \subseteq \operatorname{Int}(A) \times \operatorname{Int}(B)$, finishing the proof.