TITLE: Interior of cartesian product is cartesian product of interiors
QUESTION [11 upvotes]: I have to prove that:
$$\operatorname{Int}(A\times B) = \operatorname{Int}(A)\times \operatorname{Int}(B)$$
Where $A\subset M$ and $B\subset N$, both $M$ and $N$ metric spaces.
The problem is that the exercise does not specify the metric, so I need to try to prove it using a generic metric.
If $x\in \operatorname{Int}(A\times B)$, then an open ball can be centered at $(x,y)$ such that it is contained in $A\times B$. Therefore, this open ball, $B\big((x,y),r\big)\subset A\times B$. It means that $d\big((x,y),a\big)<r$ for any $a\in A\times B$ for some $r>0$. I guess that if I can take from here that $d(x,a)<r$ and $d(y,a)<r$ I can prove thhat $\operatorname{Int}(A\times B)\subset \operatorname{Int}(A)\times \operatorname{Int}(B)$, but I still have to prove the other way around.
Any ideas?

REPLY [11 votes]: This is true for arbitrary topological spaces (not just metrizable ones). Since $\overset{\,\,\circ}A$ is open in $M$ and $\overset{\,\circ}B$ is open in $N$, by definition $\overset{\,\,\circ}A\times \overset{\,\circ}B$ is open in $M\times N$, and clearly $\overset{\,\,\circ}A\times \overset{\,\circ}B\subset A\times B$. It follows that $$\overset{\,\,\circ}A\times \overset{\,\circ}B\subset \overset{\circ}{\overbrace{A\times B}}. $$
Conversely, since $\overset{\circ}{\overbrace{A\times B}}$ is open in $M\times N$, we may write
$$\overset{\circ}{\overbrace{A\times B}}= \bigcup_\alpha (U_\alpha\times V_\alpha) $$
where each $U_\alpha$ is open in $M$ and each $V_\alpha$ is open in $N$. Moreover, $U_\alpha\subset A$ and $V_\alpha\subset B$ for all $\alpha$, so $U_\alpha\subset \overset{\,\,\circ}A$ and $V_\alpha\subset \overset{\,\circ}B$. It follows that
$$\bigcup_\alpha (U_\alpha\times V_\alpha)\subset \overset{\,\,\circ}A\times \overset{\,\circ}B. $$