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Def: The product topology is generated by sets of the form $\prod\limits_{n\in\mathbb{N}} U_n$ where each $U_n$ is open in $X_n$ and, for all but finitely many $n$, we have $U_n = X_n$.

I am slightly confused here; does it mean that all the spaces have to be the same. I mean, if a open set in $U_3$ is to be considered for inclusion in $\prod U_n$ , $U_3$ is open in $X_3 $, and ....... $U_3$ is $X_3$ for all but finitely many n. It dont understand that last bit. I doesnt seem to make sense. How can $U_3$ and $X_3$ be rerelevant for $U_{100}$ and $X_{100}$?

user2277550
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    Obviously ‘$U_3$ is $X_3$ for all but finitely many $n$’ makes no sense; it also has nothing to do with the definition of the product topology. When we form a basic open set, $U_3$ may be any open subset of $X_3$. $U_{100}$ may be any open subset of $X_{100}$. We may pick any finite subset $F$ of $\Bbb N$ and for each $n\in F$ choose an arbitrary open $U_n\subseteq X_n$. What the definition says is that we can do this only for a finite set $F$ of indices. It can be as large as we wish, provided that it’s finite, but it cannot be infinite. And once we’ve chosen it and the open sets ... – Brian M. Scott Dec 17 '16 at 18:28
  • ... $U_n$ for $n\in F$, all of the infinitely many factors $U_n$ for $n$ not in $F$ must be the whole space $X_n$. – Brian M. Scott Dec 17 '16 at 18:28

3 Answers3

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The meaning of

  • for all but finitely many, we have $U_n = X_n$

is

  • there exists a finite subset $A \subset \mathbb{N}$ such that if $n \in \mathbb{N}-A$ then $U_n=X_n$.
Lee Mosher
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There are different kinds of product topologies.

For a family $\{X_i:i\in F\}$ of spaces:

(1). The box product topology on $\prod_{i\in F}X_i$ is generated by the base of all sets of the form $\prod_{i\in F}U_i,$ where each $U_i$ is open in $X_i.$

(2). The Tychonoff product topology on $\prod_{i\in F}X_i$ is generated by the base of all sets of the form $\prod_{i\in F}V_i$ where (i) each $V_i$ is open in $X_i,$ and (ii) the set $\{i\in F: V_i\ne X_i\}$ is finite.

The Tychonoff product topology is also called the topology of point-wise convergence. It has been a very useful tool. The box product topology is stronger, usually much stronger. If $F$ is a finite set these two topologies are equal.

(3). For some classes of spaces, e.g.metric spaces, there is a uniform product topology that is weaker than the box but stronger than the Tychonoff.

DanielWainfleet
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You have a sequence of open sets $U_1,U_2,U_3,...$; that's it, where $U_1$ is a open set of $X_1$, $U_2$ is an open set of $X_2$ and so on. i.e. $U_n$ is an open set of $X_n$ for each $n$.

And it happens that only finitely many of them are proper. For instance, suppose that only $U_1$ is strictly included in $X_1$ and $U_2$ is strictly included in $X_2$, but $U_3=X_3$, $U_4=X_4$ and so on.