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Let $A, B$ be subsets of the topological spaces $(X_1,\tau_1)\times (X_2,\tau_2)$ so so that $A\times B$ is the subset of $(X_1,\tau_1)\times (X_2,\tau_2)$.

Prove the following statements:

i) $(A\times B)'\supset A'\times B'$

ii)$\overline{A\times B}=\overline{A}\times\overline{B}$

i) Let $x,y\in (A\times B)' $. Then any open neighbourhood of $(X_1,\tau_1)\times (X_2,\tau_2)$ that contains $(x,y)$ contain some point in $A\times B\setminus\{x,y\}$. So let's take the open set $U\times V$ that contains $(x,y)$ so that $U$ contains $x$ and $V$ contains $y$. Then $U\times V\cap(A\times B)=(U\cap A)\times (V\cap B)$. As by assumption $U\times V\cap (A\times B)\neq\emptyset$ then $U\cap A\neq\emptyset $ and $V\cap B\neq\emptyset\implies x\in A' $ and $y\in B'$ so $(x,y)\in A'\times B'$. Therefore $(A\times B)'\supset A'\times B'$

ii) Using the above result it: $A\times B\cup(A\times B)'\supset A'\times B'\cup(A\times B)\implies \overline{A\times B}\supset \overline{A}\times\overline{B}$.

In order to prove the reverse inequality:

As $\overline{A}\times\overline {B}$ are the product of two sets hence closed we have:

$\overline{A\times B}\subset\overline{\overline{A}\times\overline {B}}=\overline{A}\times\overline {B}$

Note $\overline{A\times B}$ is the smallest closed set containing $A\times B$.

Therefore we have $\overline{A\times B}=\overline{A}\times \overline{B}$

Question: Is my proof right?

Thanks in advance!

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  • $\begingroup$ Fisrt, instead of $x,y\in (A\times B)^{\prime}$ you must say $(x,y)\in (A\times B)^{\prime}$. $\endgroup$ – Hector Blandin Jan 23 at 15:56

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