What is the definition of a product topology? 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?
 A: 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.
