# Interior of cartesian product is cartesian product of interiors

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 $$d\big((x,y),a\big) for any $$a\in A\times B$$ for some $$r>0$$. I guess that if I can take from here that $$d(x,a) and $$d(y,a) I can prove thhat $$\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?

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 $$\bigcup_\alpha (U_\alpha\times V_\alpha)\subset A^\circ\times B^\circ.$$
• shouldn't $A^\circ\times B^\circ$ be open in $A\times B$? Commented Apr 11, 2016 at 3:46
• Since $A^\circ\subset A$ and $B^\circ\subset B$, $A^\circ\times B^\circ\subset A\times B$ so yes, $A^\circ\times B^\circ$ is open in $A\times B$. Commented Apr 11, 2016 at 4:09