This is part of the proof of Munkres Book.
Conversely, suppose $ $ x = \left( {x_\alpha } \right) $ $ lies in the closure of , in either topology ( box or product). We show that for any given index $ $ \beta $ $ we have $ $ x_\beta \in \overline {A_\beta } $ $ . Let $ V_\beta
$ be an arbitrary open set of $ X_\beta
$ containing $ x_\beta
$ . Since $ $ \pi _\beta ^{ - 1} $ $ applied to $ V _\beta
$ is open in $ $ in either topology, it contains a point y of (this last part i don´t understand it , and sorry latex doesn´t work, but the proof is on page 116
I don´t understand why the property of being open in the product implies that exist that point, that´s my question )=