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 )=