TITLE: Closure in A Product topological space $\overline{A_1\times A_2}=\overline{A_1}\times\overline{A_2}$
QUESTION [0 upvotes]: Please is this prove is correct 
\begin{align}
(x,y)\in \overline{A_1\times A_2}&\Longleftrightarrow\forall V_X\in \mathcal{V}_X, V \cap(A_x\times A_y)\neq \emptyset\\ 
&\Longleftrightarrow \forall V_x\in \mathcal{V}_x, \forall V_2\in \mathcal{V}_y, (V_x\times V_y)\cap (A_1\times A_2)\neq \emptyset\\
&\Longleftrightarrow \forall V_x\in \mathcal{V}_x, \forall V_y\in \mathcal{V}_y, (V_x \cap A_1)\times (V_y\cap A_2)\neq \emptyset\\ 
&\Longleftrightarrow \forall V_x\in \mathcal{V}_x,\forall V_y\in \mathcal{V}_y, (V_x \cap A_1)\neq\emptyset~\text{and}~ (V_y\cap A_2)\neq\emptyset\\
&\Longleftrightarrow \forall V_x\in \mathcal{V}_x,  (V_x \cap A_1)\neq\emptyset ~\text{and}~\forall V_y\in \mathcal{V}_y (V_y\cap A_2)\neq \emptyset\\
&\Longleftrightarrow (x,y)\in \overline{A_1}\times\overline{A_2}
 \end{align}

REPLY [0 votes]: A slightly different proof might go:
if $C \subseteq X$ and $D \subseteq Y$ are closed, then $C \times D$ is closed in $X \times Y$: $$(X \times Y) \setminus (C \times D) = (X \setminus C) \times Y \cup X \times (Y \setminus D)$$ 
as $(x,y)$ not being in $C \times D$ means $x \notin C$ or $y \notin D$), and so the complement is a union of open sets in the product topology hence open.
This means that $\overline{A_1} \times \overline{A_2}$ is closed, and as it triivally contains $A_1 \times A_2$, we have $$\overline{A_1 \times A_2} \subseteq \overline{A_1} \times \overline{A_2}$$ 
To see the reverse, pick $(x,y) \in \overline{A_1} \times \overline{A_2}$ and show that $x \in \overline{A_1}$ ($y \in \overline{A_2}$ is entirely similar): let $O$ be open in $X$ containing $x$, then $O \times Y$ is product open, contains $(x,y)$ and so intersects $A_1 \times A_2$, but then $O$ intersects $A_1$ and we are done.
This argument generalises to $\overline{\prod_i A_i} = \prod_{i \in I} \overline{A_i}$ for arbitary products of spaces, making the obvious changes.
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