# Proving $(A\times B)'\supset A'\times B'$

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?

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