# Doubt about open sets in the Product topology

So we know that in the product topology a basis element is gonna be of the form $$\prod U_\alpha$$ where $$U_\alpha \neq X_\alpha$$ for a finite number of $$\alpha$$. So my thing is that we know that a product of closed sets in the product topology is gonna be closed so why is not true that if i do $$\prod X_\alpha - \prod F_\alpha = \prod X_\alpha-F_\alpha$$ im gonna get a contradiction. I know this isnt true i just cant seem to grasp it, Thanks in advance.

• We don't have $(A\setminus B)\times (C\setminus D) =(A\times C) \setminus (B\times D)$. – Berci Oct 20 at 10:07
• Yeah thats what i tought , do you know a counterexample ? – Pedro Santos Oct 20 at 10:08
• Any proper subsets yield a counterexample. – Berci Oct 20 at 10:10

Because $$\prod_\alpha X_\alpha\setminus\prod_\alpha F_\alpha\neq\prod_\alpha(X_\alpha\setminus F_\alpha)$$. In fact, $$\prod_\alpha X_\alpha\setminus\prod_\alpha F_\alpha$$ is the union of all the products $$\prod_\alpha Y_\alpha$$ where $$Y_\alpha=X_\alpha\setminus F_\alpha$$ for one specific $$\alpha$$, whereas for all other $$\alpha$$'s you have $$Y_\alpha=X_\alpha$$.

• Oh yeah thats true , Thanks. – Pedro Santos Oct 20 at 10:15

Here's a graphical demonstration: Here on the left you see $$X_1$$ (black), $$F_1$$ (red) and $$X_1\setminus F_1$$ (green). Note that I moved the latter two horizontally, so you can see all three, but actually they are all at the same line. Similarly, you see $$X_2$$, $$F_2$$ and $$X_2\setminus F_2$$ on the bottom.

In the area you see the products. The full rectangle is $$X_1\times X_2$$. The red rectangle in the middle id $$F_1\times F_2$$ Therefore $$(X_1\times X_2)\setminus(F_1\times F_2)$$ is everything of the big rectangle around the red rectangle.

On the other hand, $$(X_1\setminus F_1)\times(X_2\setminus F_2)$$ consists only of the four green rectangles.

If $$x \notin \prod_{\alpha \in J} F_\alpha$$ this means that $$x(\beta) \notin F_\beta$$ for some $$\beta \in J$$, and then the basic open set $$\prod_{\alpha \in J} O_\alpha$$ with $$O_\beta=X-F_\beta$$ and all other $$O_\alpha = X_\alpha$$ is a neighbourhood of $$x$$ that misses $$\prod_{\alpha \in J} F_\alpha$$ entirely.