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I've been trying to properly comprehend how the box topology is constructed, which I noticed I don't understand corretly when my professor asked me to show if it's false or true that the box-topology of connected spaces is connected. I know that it isn't and I've found an example in this thread here: Infinite product of connected spaces may not be connected?. The definition I'm using for the box topology is the same as from Munkres, which is:

Let $\{X_\alpha\}_{\alpha \in J}$ an indexed family of topological spaces. We will take a basis for the space $\prod_{\alpha\in J} X_\alpha$ as the set of the following products $\prod_{\alpha \in J} U_\alpha$, where $U_\alpha$ is an open set of $X_\alpha$ for every $\alpha \in J$.

Firstly, I thought that it was connected and I came with a proof: Let $A,B$ be open sets for $X = \prod_{\alpha\in J} X_\alpha$ in the box topology, where $ A = \prod_{\alpha\in J} A_\alpha$ and $B = \prod_{\alpha\in J} B_\alpha$, where $A_\alpha, B_\alpha$ are open sets for $X_\alpha$ for every $\alpha$ and $A\cup B = X$ with this union being disjoint.

Since, the union is disjoint we must have at least one $\alpha' \in J$ such that $A_{\alpha'}\cap B_{\alpha'} = \emptyset$. Since $A\cup B = X$, then $A_{\alpha'} \cup B_{\alpha'} = X_{\alpha'}$ and given the fact that $X_{\alpha'}$ is connected we gain that $X_{\alpha'} = A_{\alpha'}$ and $B_{\alpha'} = \emptyset$ without loss of generality.

So, what I did next is to say that we must have $B = \emptyset$, because it has one of its coordinates being the "empty coordinate". Otherwise, if $B$ is not the empty set, something that doesn't make sense to me, how could I pick an element that is equal to something not empty in the coordinates $B_\alpha$ for $\alpha \neq \alpha'$ while being something "empty" in the coordinate $B_{\alpha'}$?

In general, I'm really doubtful if some set similar to $B$, in the case where $B\neq \emptyset$, can be an element for the basis of the box topology. According to the definition it seems possible however from my point of view I don't see how it is viable to pick an element from such a set.

In the end, considering I wasn't so sure of what I did, I searched a little and I found the page I cite in first paragraph. My proof is obviously wrong, but I didn't managed to understand my mistake until now and I have a feeling that this is strictly related on how the box topology works somehow.

If someome can help me I'd be much obliged :)

ivantag
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  • You only covered the case that $X$ is covered by two basic open sets, not two general open sets. That makes a big difference. – Henno Brandsma May 09 '21 at 09:34

1 Answers1

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The problem with your proof is that $A$ and $B$ do not have to have the form $A = \prod_{\alpha\in J} A_\alpha$ and $B = \prod_{\alpha\in J} B_\alpha$. Remember, sets of this form are not the entire box topology, but only a basis for the box topology. So instead, $A$ and $B$ just must be unions of sets of this form.

Eric Wofsey
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  • So what you are also saying is that the following equality $\cup_{\beta \in I}\prod_{\alpha \in J} A_\alpha^\beta =\prod_{\alpha \in J} \cup_{\beta \in I}A_\alpha^\beta$ doesn't hold, right? – ivantag May 09 '21 at 21:53
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    Correct, that does not hold. I would suggest testing that out with an example where $I$ and $J$ each have just two elements and you can draw the sets as rectangles in the plane. – Eric Wofsey May 09 '21 at 21:54
  • Ok, thanks, I'll do that! However I still have a doubt: what about that type of basic open set $\prod_{\alpha \in J} B_\alpha$ where $B_\alpha' = \emptyset$ for some $\alpha'$? Does it make sense in the box topology? – ivantag May 09 '21 at 22:52
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    Sure, it's just the empty set. – Eric Wofsey May 09 '21 at 23:00
  • Even if all the others coordinates are not the empty set? – ivantag May 09 '21 at 23:29
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    Yes, a product is empty if any of its coordinates are empty. – Eric Wofsey May 10 '21 at 01:32
  • I can prove it by contradiction, right? I can suppose that it has an element in such a product, which then lead me to say that there is an element in this coordinate given by the $\emptyset$, giving us the contradiction. – ivantag May 10 '21 at 16:33
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    Sure, that works. – Eric Wofsey May 10 '21 at 20:05
  • Thanks, you helped me a lot! – ivantag May 10 '21 at 22:28