# 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 )=

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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 $\prod_\alpha A_\alpha$.

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thanks! i understand it! – Daniel Totuel Jul 11 '11 at 4:15

$\pi^{-1}(V_\beta) = \prod V_\alpha$, where $V_\alpha = X_\alpha$ if $\alpha \neq \beta$. $x$ is in this set.

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but using this , i can only guarantee that it has an $y_i$ of every index , except from beta – Daniel Totuel Jul 11 '11 at 4:13