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In a book I read the following:

Let $(E_i, \mathcal{T}_i)_{i\in I}$ be a family of topological spaces, and let $E = \prod_{i\in I} E_i$ be the Cartesian product of the $E_i$'s. Denote by $\pi_i$ the natural projection from $E$ into $E_i$, defined by $\pi_i((e_j)_{j\in I} = e_i$. The product topology on E is the topology generated by the basis consisting of the finite intersections of sets of the form $\pi^{-1}_i(X_i)$ where $X_i$ is an open set of $E_i$. These sets are nothing else than the products $\prod_{i\in I} X_i$ where the $X_i$'s are open sets of $E_i$ and where $X_i = E_i$ except for a finite number of indices.

I don't understand why the sets of finite intersections of the $\pi_i^{-1}(X_i)$ are of the form $\prod_{i\in I} X_i$ with $X_i = E_i$ cofinitely often. Because for example the set $\pi_i^{-1}(X_i)$ is in the set of finite intersection, but could well contain an element $X = \prod_{j\in I} X_j$ were all $X_j \ne E_j$ as long as $\pi_i(X) = X_i$. So why does it follows that $X_j = E_j$ infinitely often?

Martin Brandenburg
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StefanH
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2 Answers2

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$$\pi_i^{-1}(X_i)=E_1\times E_2\times ... \times X_i\times E_{i+1}\times...$$ If we intersect finitely many of these, say with indices $1..k$, for simplicity to writing up, we get $$\pi_1^{-1}(X_1)\cap \pi_2^{-1}(X_2)\cap..\cap \pi_k^{-1}(X_k)=X_1\times..\times X_k\times E_{k+1}\times E_{k+2}\times...$$ (It's also needed that, for the same index, we have $\pi_i^{-1}(X_i)\cap\pi_i^{-1}(Y_i)=\pi_i^{-1}(X_i\cap Y_i)$.)

Berci
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  • May I ask about the definition of $\pi_i^{-1}(X_i)$? I should have misunderstood it by thinking of it as

    $$\pi_i^{-1}(X_i) = {B_1\times B_2 \times ... \times X_i \times ... : B_j \in \mathcal{T}_j \text{ for all } j \in I, j\neq i}$$ i.e., the set of all cartesian products of the sets in each topology with the $i$th component equal to $X_i$. This way,

    $$\bigcap_{k=1}^n \pi_{i_k}^{-1}(X_{i_k}) = {B_1\times B_2 \times ... \times X_{i_1} ... \times X_{i_k} \times ... : B_j \in \mathcal{T}_j \text{ for all } j \in I, j\neq i_1...i_k}$$

     – sam2018 Jul 13 '23 at 16:55
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Think about what each of the generating elements $\pi_i^{-1} (X_i)$ look like in the case each of your $E_i = \Bbb R$ and $X_i$ is an open interval. Your open sets look like things of the form $$ \Bbb R \times \Bbb R \times \dots \times (a,b) \times \dots \times \Bbb R \times \Bbb R \times \dots$$

Finite intersections of such sets will only ever get you a finite number of intervals anywhere but the majority of the sets in the product will still be $\Bbb R$.

muzzlator
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  • if $X_1 = (0,1)$ and $X = (0,1) \times (1,2) \times (2,3) \times \ldots \in \pi_1^{-1}(X_1)$, then $\pi_1(X) = (0,1)$, but none of the individual components equals $\mathbb{R}$. – StefanH Apr 07 '13 at 14:34
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    We are not defining the open sets in the product topology as simply things whose $\pi_i$ is $X_i$. Rather we're taking $\pi_i^{-1}(X_i)$ and intersecting them. – muzzlator Apr 07 '13 at 14:36
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    ahh, yes I understand! – StefanH Apr 07 '13 at 14:39