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Let $X,Y$ be topological spaces, $A \subset X, B \subset Y$. Show that $\overline{A \times B} = \bar{A} \times \bar{B}$.

Well obviously this is a question from Munkres, following is my attempt, and please point out the mistake if there is any.

I intend to tackle this question from the definition of closure, that is closure of $A$, $\bar{A}$ is defined as all the intersection of closed sets containing $A$.

Let $C \subset X$ such that $C$ closed (in $X$)., $A \subset C$. Similarly let $D \subset Y$, $D$ closed (in $Y$) such that $B \subset D$.

All the closed subsets containing $A \times B$ would be of the form $C \times D$ (this is the part I think could be wrong though, since I am assuming all the closed sets containing $A \times B$ are of the form $C \times D$, even if I verified that $C \times D$ is indeed closed set in the product topology.

It follows that $\overline{A \times B} = \bigcap_{C \supset A, D \supset B} C \times D = \bigcap_{C \supset A} C \times \bigcap_{D \supset B} D = \bar{A} \times \bar{B} $

Thank you!

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1 Answer 1

up vote 1 down vote accepted

Your concern is justified: it’s not in general true that all closed subsets of $X\times Y$ containing $A\times B$ have the form $C\times D$. For instance, take $X=Y=\Bbb R$, $A=B=[0,1]$; then the closed ball of radius $2$ centred at the origin, $\{\langle x,y\rangle\in\Bbb R^2:x^2+y^2\le 4\}$, is a closed set in $\Bbb R^2$ that contains $A\times B$ and is not the Cartesian product of two sets.

This is a result for which the characterization of $\operatorname{cl}A$ as the intersection of all closed sets containing $A$ is not the best choice; you’re better off combining the characterization that $x\in\operatorname{cl}A$ iff every open nbhd of $x$ meets $A$ with the definition of the product topology of $X\times Y$.

The intersection characterization isn’t wholly useless here: once you show that $\operatorname{cl}A\times\operatorname{cl}B$ is closed in $X\times Y$, you can use it to see immediately that $\operatorname{cl}(A\times B)\subseteq\operatorname{cl}A\times\operatorname{cl}B$, but it’s not very helpful for the opposite inclusion. It’s much easier simply to let $\langle x,y\rangle\in\operatorname{cl}A\times\operatorname{cl}B$ and show that every open nbhd of $\langle x,y\rangle$ meets $A\times B$.

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Thanks a lot for the help in justification! –  Daniel Feb 13 '13 at 2:27
    
@jsk: My pleasure! –  Brian M. Scott Feb 13 '13 at 2:27

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