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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}

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    shouldn't the first $\mathcal{V}_X$ and $V_X$ be without $X$? then look at the equivalence from line 1 to line 2: it is not true that each $V\in\mathcal{V}$ has the form $V=V_X\times V_Y$ with $V_X\in\mathcal{V}_X$ and $V_Y\in\mathcal{V}_Y$. So here is the crucial point in your proof. Find a reason why it is enough to show that for sets of that special form as a product. – 2017-01-16
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    I must work with open set ? i mean with $X=(x,y)$ – 2017-01-16
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    @Max how i cn correct this ? please – 2017-01-16
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    If the main purpose of your post is to check whether your proof is correct (as opposed to asking for any proof of this fact), you should use ([tag:proof-verification]) to make this clear. – 2017-01-17

2 Answers 2

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A minor quible as was already commented: you go from "every open neighbourhood $V$ of $(x,y)$ intersects $A_1 \times A_2$ to "for every neighbourhood $V_x$ of $x$ and every neighbourhood $V_y$ of $y$, $(V_x \times V_y)$ intersects $ A_1 \times A_2$". This is not a purely logical deduction (as you seem to imply ), but the consequence of a simple topological observation:

Fact: Let $X$ be a topological space with a base $\mathcal{B}$, then for every $x \in X$ and every $A \subseteq X$: $x \in \overline{A}$ iff for $\forall B \in \mathcal{B}: (x \in B) \rightarrow B \cap A \neq \emptyset$.

left to right is trivial, as base elements are just open sets, but for right to left: let $O$ be open in $X$ with $X \in O$. The as $\mathcal{B}$ is a base, there is some $B \in \mathcal{B}$ with $x \in B \subseteq O$. By the right side condition, $A \cap B \neq \emptyset$ so a fortiori: $A \cap O \neq \emptyset$, so $x \in \overline{A}$, as $O$ was arbitrary.

So closure testing can be done with base elements only. The above fact is used implicitly in a lot of proofs. It also suffices to test inverse images of base sets of continuity etc.

So you could remark that this biimplication does hold, because sets of the form $U \times V$ form a base for the product topology.

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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.