About the proof of $\overline{\prod A_{\alpha}}\subset \prod\overline A_{\alpha}$

Question

I found this proof here products of closure, is the closure of the product

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 $\prod_\alpha A_\alpha$.

I'm having problem understanding the bold text and also I don't see the conclusion of the proof, I mean the ...therefore "for any given index $\beta$ we have $x_\beta \in \overline{A_\beta}$"

Could anyone help me, please ?

Answer 1

You’re assuming that $x\in\overline{\prod_{\alpha}A_{\alpha}}$ and want to prove that $x_{\beta}\in\overline{A_{\beta}}$ for each $\beta$.

Let $V_{\beta}$ be an open subset of $X_{\beta}$ that contains $x_{\beta}$. Letting $$V\equiv \pi_{\beta}^{-1}(V_{\beta})=V_{\beta}\times\prod_{\alpha\neq\beta}X_{\alpha},$$ it is easy to see that $V$ is an open subset of $\prod_{\alpha}X_{\alpha}$ (in either the product or the box topology). Observe also that $x_{\beta}\in V_{\beta}$ implies that $x\in V$. But since $x$ is also contained in the closure of $\prod_{\alpha}A_{\alpha}$ and $V$ is open, there must exist some $y\in V\cap \prod_{\alpha}A_{\alpha}$. This implies, in particular, that $y_{\beta}\in V_{\beta}\cap A_{\beta}$.

To summarize, what has been shown is that each open set containing $x_{\beta}$ also contains an element of $A_{\beta}$, which implies that $x_{\beta}$ is in the closure of $A_{\beta}$, as claimed.

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Throughout the proof, we use the following fact:

Let $X$ be an arbitrary topological space, $A\subseteq X$, and $x\in X$. Then, $x\in\overline{A}$ if and only if every open set $U\subseteq X$ that contains $x$ also contains an element from $A$.

Answer 2

1) $\pi_{\beta}^{-1}(V_{\beta})=V_{\beta}\times\displaystyle\prod_{\alpha\ne\beta}X_{\alpha}$ and the latter is an open set in either box or product topology. Now $x\in V_{\beta}\times\displaystyle\prod_{\alpha\ne\beta}X_{\alpha}$, so $\left(V_{\beta}\times\displaystyle\prod_{\alpha\ne\beta}X_{\alpha}\right)\bigcap\displaystyle\prod_{\alpha}A_{\alpha}\ne\emptyset$, and hence $\pi_{\beta}^{-1}(V_{\beta})\bigcap\displaystyle\prod_{\alpha}A_{\alpha}\ne\emptyset$.