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

Question

Let $\{X_{\alpha}\}$ be an indexed family of spaces; let $A_\alpha\subset X_\alpha$ for each $\alpha$. If $\prod X_\alpha$ is given either the product or the box topology, then $$\prod\overline A_{\alpha}=\overline{\prod A_{\alpha}}$$

Proof. This $\subset$ side

Let $x=(x_\alpha)\in\prod\overline A_{\alpha}.$ Let $U=\prod U_\alpha$ be a basis elements for either the box or product topology that contains $x.$

Since $x\in\overline A_\alpha,$ we can choose a point $y_\alpha\in U_\alpha\cap A_\alpha$ for each $\alpha$.

Then $y=(y_\alpha)$ belongs to both $U$ and $\prod A_\alpha.$

Since $U$ is arbitrary, it follows that $x$ belongs to $\overline{\prod A_{\alpha}}$.

I don't understand the bold text, could anyone explain this please?

Answer 1

The crux is the following characterisation of the closure of $A$ in a space $X$:

$x \in \overline{A}$ iff for every (basic) open subset $O$ of $X$ that contains $x$ we have that $O \cap A \neq \emptyset$.

The first bold statement holds as $x_\alpha \in \overline{A_\alpha}$ and $x_\alpha \in U_\alpha$ and the $U_\alpha$ is open in $X_\alpha$. (we apply the left to right direction here.)

The last bold statement follows from the right to left implication applied in the product. We started with an arbitary basic open subset of the product that contains $x$ and showed it intersects $A$.