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 December 29th, 2017, 02:06 PM #1 Newbie   Joined: Dec 2017 From: vienna Posts: 12 Thanks: 1 Product topology why is (-1/n ,1/n) not an open in the product topology but open in the box topology? Thanks from greg1313
 December 29th, 2017, 02:09 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,157 Thanks: 631 You mean the open interval? I don't see how you're relating this to the product topology. It looks like you're in $\mathbb R$ there. Can you clarify please?
 December 29th, 2017, 05:02 PM #3 Newbie   Joined: Dec 2017 From: vienna Posts: 12 Thanks: 1 ah i m sorry the product of (-1/n,1/n) and n goes from 1 to infinity and yeah in R
December 29th, 2017, 05:16 PM   #4
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 Originally Posted by aminea95 ah i m sorry the product of (-1/n,1/n) and n goes from 1 to infinity and yeah in R
Yes thanks this is an interesting problem. The box topology is the topology generated by the Cartesian products of all the open sets in the component spaces.

The product topology is the weakest topology that makes all the projection maps continuous. By weakest, it's the topology that's a subset of every other possible topology. We often say it's the topology with the fewest open sets, where one topology has fewer open sets than another if the first one's a proper subset of the second. Nothing to do with cardinality.

The box topology and the product topology are the same in the case of a finite Cartesian product but they differ in the infinite case. The product of the intervals (-1/n, 1/n) is obviously open in the box topology. So it will be interesting to work out why it's not open in the product topology.

 December 29th, 2017, 08:22 PM #5 Senior Member   Joined: Aug 2012 Posts: 2,157 Thanks: 631 Ok I have the outline but I'm missing a couple of key details. Let $\displaystyle X = \Pi_\infty \ \mathbb R$ where the notation indicates countably many copies of $\mathbb R$. Each projection is a map $\pi_n : X \to \mathbb R$ given by $\pi_n(x) = x_n$, where $x_n$ is the $n$-th coordinate of $x$. Saying that a projection is continuous means that the inverse image of any open set is open. Each $I_n = (\frac{-1}{n}, \frac{1}{n})$ is open in the usual topology on $R$, so that its inverse image is (*) $\pi_n^{-1} (I_n)= \Pi \ \mathbb R \times \mathbb R \times \dots \mathbb R \times I_n \times \mathbb R \times \mathbb R \dots$ In other words all the components are $\mathbb R$ except for the $n$-th component, which is $I_n$. Each inverse image $\pi_n^{-1} (I_n)$ is open by definition of the product topology (which makes all the projections continuous) and moreover there are no other sets in the topology that can't be made from these inverse images. All the other necessarily open sets are the inverse images under $\pi_n$ of all the open open sets in $I_n$, and for the moment we'll ignore these. Now the product topology is the smallest topology that makes all the projections continuous. So the product topology consists of exactly the inverse images (*) along with all their arbitrary unions of finite intersections. In other words the product topology on $X$ is the topology generated by the subbase $\{\pi_n^{-1} (I_n)\}_{n \in \mathbb N}$ See https://en.wikipedia.org/wiki/Subbase for background. Missing detail #1: I claim that if we threw in all the inverse images of the open subsets of the $I_n$'s we wouldn't get any new open sets in the product topology that will make any difference to the final conclusion. This needs proof. Now a moment's thought ["proof by a moment's thought ..."] shows that the any finite intersection of these subbase elements is a product where all the components are the reals except for finitely many components that are $I_n$'s. And if you take arbitrary unions of these finite intersections, then it's clear that $\Pi \ I_n$ is not any open set in this topology. Missing detail #2: Arbitrary unions of finite intersections of subbase elements can't generate an infinite product of the $I_n$'s in the obvious way, but how about in some not-obvious way? That needs proof. OP is this enough to work with? What do you think? If I get a chance I'll see if I can fill in the missing details and remove some of the fuzziness about the subbase business, but most likely I'll just leave that up to you or anyone else who feels like finishing this off. My ignoring the open subsets of the $I_n$'s is not explained well. The product topology is generated by all of the inverse images of the $I_n$'s and all of their respective open subsets. Thanks from greg1313 and Ould Youbba Last edited by Maschke; December 29th, 2017 at 08:43 PM.
 December 30th, 2017, 08:58 AM #6 Member   Joined: Jan 2016 From: Athens, OH Posts: 92 Thanks: 47 The Wikipedia article https://en.wikipedia.org/wiki/Product_topology confirms my memory of the definition of the product topology: Given $\{X_i\,:\,i\in I\}$, a set of topological spaces, $\prod_{i\in I}U_i$ is open in $\prod_{i\in I}X_i$ if and only if $U_i$ is open in $X_i$ and only finitely many of the $U_i$ are unequal to $X_i$. So it is now obvious that the infinite product of open intervals unequal to R is not open; furthermore such a product has an empty interior!
December 31st, 2017, 08:39 AM   #7
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 Originally Posted by johng40 The Wikipedia article https://en.wikipedia.org/wiki/Product_topology confirms my memory of the definition of the product topology: Given $\{X_i\,:\,i\in I\}$, a set of topological spaces, $\prod_{i\in I}U_i$ is open in $\prod_{i\in I}X_i$ if and only if $U_i$ is open in $X_i$ and only finitely many of the $U_i$ are unequal to $X_i$. So it is now obvious that the infinite product of open intervals unequal to R is not open; furthermore such a product has an empty interior!
Yep, and this is proven in a quite straightforward way. Assume that $\prod_{i\in I} U_i$ has a nonempty interior, say that it contains $(x_i)_{i\in I}$. A basis of the product topology is given by $\prod_{i\in I} V_i$ where only finitely many of the $V_i$ equal $X_i$. Thus by definition of a basis, we have some basis set such that
$$(x_i)_{i\in I} \in \prod_{i\in I} V_i\subseteq \prod_{i\in I} U_i.$$
But some reflection shows immediately that since $V_i = X_i$ for all but finitely many $i$, we also have $U_i = X_i$ for all but finitely $i$.

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