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Please consider the statement and it's proof below:

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The author further says that : Members of the product topology can all be expressed as union of products, but most members of the product topology are not products.

For example - the open disc $\{x \in \mathbb R^2~|~x_1^2+x_2^2 <1 \}$ is a member of the product topology on $\mathbb R^2$ but it is not a product of open intervals of $\mathbb R$. Infact, to express it as a union of products of open intervals, we need to use an infinite collection of such products in the following way: For each $x \in U$, where $U$ is an open set in the disc, set $r_x=1- \sqrt{x_1^2+x_2^2}$. Then $U$ can be expressed as the union $\bigcup \{(x_1-r_x~,~ x_1+r_x) \times (x_2-r_x~,~ x_2+r_x)~|~x \in U\}$

Is there a contradiction between the second para of theorem $4.5.1$ and the example cited above?

According to the second para, $W$ any open subset can be expressed as the union of products of open balls ( intervals here in case of $\mathbb R$ ). Then, the author refutes that this is not possible in the example.

I am confused, any help will be greatly appreciated. Thanks!

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  • $\begingroup$ Where do you see a contradiction? $\endgroup$ – José Carlos Santos May 10 at 8:33
  • $\begingroup$ According to the second para, $W$ any open subset can be expressed as the union of products of open balls ( intervals here in case of $\mathbb R$ ). Then, the author refutes that this is not possible in the example. $\endgroup$ – MathMan May 10 at 8:35
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    $\begingroup$ The theorem says you have to take unions of rectangles. The example shows how to express the disk as a union of rectangles. So everything is OK. $\endgroup$ – Michal Adamaszek May 10 at 8:36
  • $\begingroup$ @MichalAdamaszek The example says we will have to take an infinite union of rectanges. $\endgroup$ – MathMan May 10 at 8:38
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    $\begingroup$ @MathMan Yes, of course. The theorem does not restrict the union to be finite. It would not make any sense, since the topology axioms anyway say that any union of open sets is still open. $\endgroup$ – Michal Adamaszek May 10 at 8:39
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There is no contradiction. In that example, the author explains that:

  • the open disk is an open set with respect to the product topology;
  • it is not a product of two open intervals;
  • it is an union of products of open intervals.

And all of this is consistent with the theorem that the author proved.

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  • $\begingroup$ Thank you very much for the answer. I have some questions still in mind. If I could please ask. The first is, the author says : $1$. The open disc is a member of the product topology. Is that because the euclidean metric is a conserving metric and thus, any open set as described in the theorem is a member of the product topology? $\endgroup$ – MathMan May 10 at 8:42
  • $\begingroup$ I don't know what a conserving metric is. Anyway, the open disk is open in the product topology because, as the author explains in that example, it can be written as an union of products of open subsets of $\mathbb R$ (which happen to be open intervals). $\endgroup$ – José Carlos Santos May 10 at 8:45
  • $\begingroup$ okay. Could you please clarify why does the author say then that most members of the product topology are not products? $\endgroup$ – MathMan May 10 at 8:48
  • $\begingroup$ Because most open subsets of $\mathbb R^2$ are not products of open subsets of$\mathbb R$. Besides the open disks you have, for instance, $\mathbb R^2\setminus\{p\}$, for some $p\in\mathbb R^2$, or $\{(x,y)\in\mathbb R^2\,|\,x\neq y\}$. $\endgroup$ – José Carlos Santos May 10 at 8:51
  • $\begingroup$ But, $\bigcup \{(x_1-r_x~,~ x_1+r_x) \times (x_2-r_x~,~ x_2+r_x)~|~x \in U\}$ is the union of product of open intervals of $\mathbb R$ ? $\endgroup$ – MathMan May 10 at 8:54

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