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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.

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

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  • $\begingroup$ Oh yeah thats true , Thanks. $\endgroup$ – Pedro Santos Oct 20 at 10:15
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Here's a graphical demonstration:

Image of product of differences

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.

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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.

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