Interior of cartesian product is cartesian product of interiors

I have to prove that:

$$Int(A\times B) = Int(A)\times Int(B)$$

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 $x\in Int(A\times B)$, then an open ball can be centered at $(x,y)$ such that it is contained in $A\times B$. Therefore, this open ball, $B((x,y),r)\subset A\times B$. It means that $d((x,y),a)<r$ 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 $Int(A\times B)\subset Int(A)\times Int(B)$, but I still have to prove the other way around.

Any ideas?

Solution 1:

This is true for arbitrary topological spaces (not just metrizable ones). Since $A^\circ$ is open in $M$ and $B^\circ$ is open in $N$, by definition $A^\circ\times B^\circ$ is open in $M\times N$, and clearly $A^\circ\times B^\circ\subset A\times B$. It follows that $$A^\circ\times B^\circ\subset (A\times B)^\circ. $$

Conversely, since $(A\times B)^\circ$ is open in $M\times N$, we may write $$(A\times B)^\circ = \bigcup_\alpha (U_\alpha\times V_\alpha) $$ 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 $U_\alpha\subset A^\circ$ and $V_\alpha\subset B^\circ$. It follows that $$\bigcup_\alpha (U_\alpha\times V_\alpha)\subset A^\circ\times B^\circ. $$