products of closure, is the closure of the product 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 )=
 A: $\pi^{-1}(V_\beta) = \prod V_\alpha$,  where  $V_\alpha = X_\alpha$  if  $\alpha \neq  \beta$.  $x$  is in this set.
A: Filling in the blanks in the question, I think that it should read something like this:
Conversely, suppose that $x = (x_\alpha)$ lies in the closure of $\prod_\alpha A_\alpha$ in either the box or the product topology on $\prod_\alpha X_\alpha$. 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 either topology, it contains a point $y$ of $\prod_\alpha A_\alpha$.
By definition $\pi_\beta^{-1}[V_\beta] = \{y = (y_\alpha) \in \prod_{\alpha}X_\alpha:y_\beta \in V_\beta\}$; as Stephen said, this is simply $\prod_\alpha V_\alpha$, where $V_\alpha = X_\alpha$ if $\alpha \ne \beta$. This set is open in both the box and the product topologies on $\prod_\alpha X_\alpha$, and it contains $x \in \overline{\prod_\alpha A_\alpha}$ (since $x_\beta \in V_\beta$), so it must contain some point $y \in \prod_\alpha A_\alpha$: the fact that $x \in \overline{\prod_\alpha A_\alpha}$ means by definition that every open nbhd of $x$ must contain some point of $\prod_\alpha A_\alpha$.
