In a product topology $X\times Y$, the open sets are defined as $V\times W$, where $V\in X$ and $W\in Y$ are open sets.

If $X\times Y$ is a topology, then the union of open sets should be open- i.e. $(A\times B)\cup (C\times D)=M\times N$ for some open sets $M\in X$ and $N\in Y$. We know that $(A\times B)\cup (C\times D)\neq (A\cup C)\times (B\cup D)$. How do we prove that the open sets $M$ and $N$ exist?

Thanks in advance!

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    $\begingroup$ No, that is not the definition of the open sets in the product topology. $\endgroup$ – Thomas Andrews Aug 29 '13 at 15:36
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    $\begingroup$ Indeed, this is just a basis for the product topology. $\endgroup$ – Agustí Roig Aug 29 '13 at 15:37
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    $\begingroup$ So you just have to show that the intersection of two boxes is a union of boxes in order to show that this is a basis. $\endgroup$ – Stefan Hamcke Aug 29 '13 at 15:38
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    $\begingroup$ Yes, but not all open sets are base sets. $\endgroup$ – Stefan Hamcke Aug 29 '13 at 15:39
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    $\begingroup$ In order to show you the difference between a basis and all open sets: $(0,1)$ and $(2,3)$ are open sets of a basis of the usual topology of $\mathbb{R}$, as all open intervals are, right? But not every open set of $\mathbb{R}$ is an interval: $(0,1) \cup (2,3)$ is an open set, but not an interval. So you can not expect that all open sets in the product topology are supposed to be just "boxes" like $M \times N$. Another way to say it: a union of open "boxes" is an open set, but doesn't need to be a "box", necessarily. $\endgroup$ – Agustí Roig Aug 29 '13 at 15:46

The definition of the product topology is not as you've defined it above. Those sets are not closed under union, as you've noted.

The product topology defines $S\subset X\times Y$ as open if $S$ can be represented as the union of sets $U\times V$ where $U$ is an open subset of $X$ and $V$ is an open subset of $Y$.

So the sets of the form $U\times V$ with $U,V$ open in $X,Y$, respectively, form a basis of the product topology, not the entire topology.

  • $\begingroup$ @ThomasAndrews- So sets of the form $U\times V$ are not base sets? Because base sets are open, and so are their unions. $\endgroup$ – fierydemon Aug 29 '13 at 16:37
  • $\begingroup$ No, they are a basis. They are not the entire topology. $\endgroup$ – Thomas Andrews Aug 29 '13 at 16:55
  • $\begingroup$ Notice my definition: The open sets in the product topology are any union of the sets of the form $U\times V$. So in particular, $(U_1\times V_1)\cup(U_2\times V_2)$ is of this form, so it is open. $\endgroup$ – Thomas Andrews Aug 29 '13 at 16:58
  • $\begingroup$ OK I suppose I got it now. $U\times V$ is open, and so is $(U_1\times V_1)\cup (U_2\times V_2)$. However, not every open set can be represented this way. There are open sets which are unions of sets of this form. $\endgroup$ – fierydemon Aug 29 '13 at 17:04
  • $\begingroup$ Well, every open set is some union of sets of this form - possibly infinite sets. Try thinking of the case where $X=Y=\mathbb R$ with the usual topology. Then the product topology on $\mathbb R\times R$ is the same as the usual topology on $\mathbb R^2$. So try to represent the usual unit disk $\{(x,y):x^2+y^2<1\}$ as the union of open rectangles. Hint: you'll need infinitely many rectangles to get the whole open disk. $\endgroup$ – Thomas Andrews Aug 29 '13 at 17:08

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