Let $(E_i, \mathcal{T_i})$ a family of topological spaces and $E=\prod_{i \in I} E_i$.
The product topology is the smallest topology on $E$ such that all projection maps are continuous.
the base of this topology is given by :
$\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$ for all $i\in I$, and $U_i =E_i$ for all but finitely many $i\in I$}\}$.
The problem is why we define the base of the product topology like that and didn't define it as in the finite case : $\mathcal B=\{ \prod_{i\in I} U_i : \text{ $U_i$ open in $E_i$ for all $i\in I$} \}$.
I know that $\mathcal B$ in this case is the base of the box topology but I didn't understand yet.
Thank you in advance.
Best Answer
Hint
If you have a collection of set $\mathcal A=\{A_i\}_{i\in\mathcal I}$, the coarse topology that contains $\mathcal A$ is the topology of basis $$\mathcal B=\left\{\bigcap_{j\in J}A_j\mid |\mathcal J|<\infty ,A_j\in \mathcal A\right\}.$$
In your case, denote $\pi_k: \prod_{\ell\in \mathcal K}E_\ell\to E_k$ defined by $$\pi_k((x_\ell)_{\ell\in\mathcal K})=x_k.$$ In particular, $\pi_k$ is continuous if and only if $$\pi_{k}^{-1}(U)=E_1\times ...\times E_{k-1}\times U\times E_{k+1}\times ...$$ is open in $\prod_{\ell\in \mathcal K}E_\ell$ for all $U$ open in $E_k$. So, in you case, $\mathcal A$ is a collection of sets of the form $\prod_{\ell\in \mathcal K}V_\ell$, where $V_\ell$ are open in $E_\ell$ and $V_\ell=E_\ell$ for all $\ell\in \mathcal K$, but one. Then $\mathcal B$ is the collection of sets of the form $\prod_{\ell\in\mathcal K}V_\ell$ where $V_\ell$ are open, and $V_\ell=E_\ell$ for all $\ell$, but finitely many $\ell$'s.