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Let $A$ be a subset of the topological space $X$ and let $B$ be a subset of the topological space $Y$. Show that in the space $X \times Y$, $\overline{(A \times B)} = \bar{A} \times \bar{B}$.

Can someone explain the proof in detail? The book I have kind of skims through the proof and I don't really get it.

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  • $\begingroup$ Which details you don't understand about your book's proof $\endgroup$ – Amr Feb 13 '13 at 15:25
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    $\begingroup$ Use \overline instead of \bar: \overline{A \times B} gives $\overline{A \times B}$ $\endgroup$ – Hurkyl Feb 13 '13 at 15:27
  • $\begingroup$ @Hurkyl thanks I was trying to figure that out $\endgroup$ – user39794 Feb 13 '13 at 15:28
  • $\begingroup$ @Amr well the proof I'm reading says: Let (x,y) in closure of (AxB). They there is a basis elt UxV such that U is open in X and V is open in Y, and such that (x,y) is in UxV and (UxV) intersect (AxB) is nonempty. It follows that (U intersect A) x (V intersect B) is nonempty. So (x,y) is in closure of A x closure of B. .........I don't understand how the second to last sentence implies the last sentence. And that's just the first direction. I don't understand the second direction at all. $\endgroup$ – user39794 Feb 13 '13 at 15:29
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A set in $X\times Y$ is open iff it is the union of sets of type $U\times V$ with $U,V$ open.

Therefore, if $(x,y)\notin\overline{A\times B}$, there is $U\times V$ containing it and disjoint to $A\times B$. As $(U\cap A)\times (V\cap B)\subseteq (U\times V)\cap (A\times B)$, one of the sets $U\cap A$ or $V\cap B$ is empty, i.e. $x\notin \overline A$ or $y\notin \overline B$. This shows $\overline A\times\overline B \subseteq \overline{A\times B}$. The converse is not much different.

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    $\begingroup$ Actually, the converse can be much simpler, as shown in my answer. $\endgroup$ – Herng Yi Feb 13 '13 at 15:49
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$(\subseteq)$: The product of closed sets $\overline{A} \times \overline{B}$ is closed. For every closed $C$ that contains $\overline{A} \times \overline{B}$, $A \times B \subseteq C$ so $\overline{A \times B} \subseteq \overline{\overline{A} \times \overline{B}} = \overline{A} \times \overline{B}$.

$(\supseteq)$: Choose any $(a,b) \in \overline{A} \times \overline{B}$. Notice that for every open neighborhood $W \subseteq X \times Y$ that contains $(a, b)$, $U \times V \subseteq W$ (by the definition of the product topology) for some open neighborhood $U$ of $a$ and some open neighborhood $V$ of $b$. By the definition of closure points, $U$ intersects $A$ at some $a'$. Similarly, define $b' \in V \cap B$. Hence, $(a', b') \in W \cap (A \times B)$.

To summarize, every open neighborhood $W \subseteq X \times Y$ that contains $(a,b)$ must intersect $A \times B$, therefore $(a,b) \in \overline{A\times B}$.

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  • $\begingroup$ haha no problem, seems like you were in the right direction! $\endgroup$ – Herng Yi Feb 13 '13 at 15:55
  • $\begingroup$ Could you explain how showing $(a', b') \in W \cap \overline{A \times B}$ shows $\overline{A} \times \overline{B} \subseteq \overline{A \times B}?$ Because you assumed $(a,b) \in \overline{A} \times \overline{B},$ so shouldn't you show that $(a,b) \in \overline{A \times B}?$ $\endgroup$ – dhk628 Oct 14 '15 at 14:02
  • $\begingroup$ I've elaborated on my answer; $W$ intersects $A \times B$ for all open neighborhoods $W$ of $(a,b)$, so $(a,b)$ is a point of closure. $\endgroup$ – Herng Yi Oct 17 '15 at 2:52

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