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I am new to advanced mathematics and I recently started reading a book on topology. I am struggling to understand what it is saying in this paragraph. This is what it says:

Let $E_i$ $(i=1,2,...,n)$ be a finite family of topological spaces and let $E=\Pi E_i(i=1,2,...,n)$ be the collection of sequences $(x=x_1, x_2,...,n)$ where $x_i \in E_i$. Among all the possible topologies on $E$ we shall take only those for which each of the projections $x \rightarrow x_i = f_i(\omega_i)$ is continuous. This condition amounts to saying that for every open set $\omega_i \subset E_i$, the set $f_i^{-1}(\omega_i)$, which is simply the product

$\qquad\qquad\qquad\qquad E_1\ \times\ ... \times\ E_{i-1}\ \times\ \omega_i\ \times\ E_{i+1}\ \times\ ... \times\ E_n,$

should be open in $E$.

I understand the definition of a product topology given by Wolfram MathWorld, but this notation used in this definition is confusing me to some degree. Can anyone explain what this definition is saying?

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    $\begingroup$ That's not a definition as they do not specify which topology is used from all those topologies which make the projections continuous. In fact we define the product topology as the coarsest of these topologies. There is another one, called the box topology, but for finite products they agree, only for infinite products they are different. $\endgroup$ – Stefan Hamcke Mar 12 '14 at 19:16
  • $\begingroup$ For finite number of sets the two notions are equivalent,take the intersection of products given above and u can get the product of open sets in the individual spaces. $\endgroup$ – viplov_jain Mar 12 '14 at 19:18
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The point of the product topology is the following. Given a collection of topological space $\{(E_i, T_i)\}_{i \in I}$ (for some index set $I$ which may be finite, may be infinite) we have natural maps

$$ p_k : \mathbb{E} = \times_{i \in I} E_i \to E_k $$

for each $k \in I$, where $\times_{i \in I} E_i$ is the usual set-wise cartesian product of the underlying sets $E_i$. It is natural to ask that these functions be continuous, and so whatever topology we put on $\mathbb{E}$ ought to have that property.

Well, what would this require? A continuous map is such that $p_k^{-1}(U)$ must be open for every open set $U$. In our case, we would have a minimal requirement that the sets

$$ p_k^{-1}(U) = U \times \Big(\times_{i \in I, k \neq i} E_i\Big) $$

are open. So let us choose these sets as a sub-basis for our topology. That is, the topology on $\mathbb{E}$ will consist of these sets as well as all finite intersections of them. Or, more explicitely, it will consist of sets of the form

$$ \times_{i \in I} U_i $$

with $U_i$ open in $E_i$, and such that all but finitely many $U_i$ are equal to $E_i$. For finite sets, of course, this is just sets of the form

$$ U_1 \times \cdots \times U_m $$

The point then is that any topology for which the maps $p_k$ are all continuous will be finer than this topology i.e. they will have more open sets; that is, this is the coarsest such topology which ensures that all of these maps are continuous.

Short form: all we are asking is for projections to be continuous.

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    $\begingroup$ I should comment that my indexing in $p_k^{-1}(U)$ is a bit off, but I'm too lazy to think of a better way to write that. If it helps, pretend $I$ is the natural numbers, in which case we could write $p_k^{-1}(U) = E_1 \times \cdots \times E_{k-1} \times U \times E_{k+1} \times \cdots$. $\endgroup$ – Simon Rose Mar 12 '14 at 19:45

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