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Show that every open set $A$ is in a metric space $(X,d)$ is the union of closed sets.

This is a question on my analysis homework. I understand that this can only be true if we consider the union of infinite closed sets. However, I am not sure what I can do. I understand to prove a set is open, then a ball centered at an arbitrary point with a radius will be completely contained in the set. But what is the radius? Is this the correct approach?

Empty
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john doe
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  • Possible duplicate of http://math.stackexchange.com/questions/498242/open-set-is-a-union-of-an-infinite-number-of-closed-sets – lhf Sep 19 '13 at 06:00
  • The countable union of closed sets is called a $F_\sigma$ (the $F$ from the french fermé=closed). – Mauricio Tec Sep 19 '13 at 06:01

3 Answers3

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In fact, every subset of a metric space is the union of closed sets.

Hint: In a metric space singleton sets $\{x\}$ are closed.

Hagen von Eitzen
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Hint: It is equivalent to "Every closed set is the intersection of open sets". Let $C$ is a closed set. Consider $$N(C,r)=\{x\in X: d(x,y)<r \text{ for some }y\in C\}.$$ We can show that $\bigcap_{r>0} N(C,r)$ is equal to the closure of $C$

Hanul Jeon
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  • I'm not getting your point. You have written that every closed set is the intersection of open sets. But finally you have shown than $C$ is the union of open sets. Please clarify it. – Empty Jul 03 '17 at 07:40
  • @s717717 Every open set is a complement of some closed set, and vice versa. So we can take the complement to sets in the original problem – Hanul Jeon Jul 03 '17 at 08:11
  • Yes..But you have taken here $C$ as a closed set – Empty Jul 03 '17 at 08:33
  • @s717717 I've found my answer has a typo. Is my answer fine now? – Hanul Jeon Jul 03 '17 at 08:37
  • Yes. But how can I show that $\bigcap_{r>0} N(C,r)=\bar C$ ? In rough sketching I'm not getting that. Can you give some hint to prove this..? – Empty Jul 03 '17 at 19:53
  • @S717717 We assume that $C$ is closed, so the reversed inclusion holds. To prove other direction, check that if $x\in \bigcap_{r>0} N(C,r)$ then for any $r>0$, $B(x,r)\cap C$ is not empty. – Hanul Jeon Jul 03 '17 at 20:30
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Since $A$ is open it is the union of open balls, call them $\{B_\alpha\}$. Then try to prove that each $B_\alpha$ is the union of closed sets, then since $A=\bigcup B_{\alpha}$ you would have your result.

Sketch: Hence if we have a ball around $x$ we have that $B_{\alpha}=\{y\in X: d(x,y)<\epsilon\}$ for some $\epsilon$. Write down closed balls of radius smaller than $\epsilon$, but which radii increase towards $\epsilon$.

Daniel Montealegre
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  • Let me see if understand. If A is an open set, then there is an open ball with an arbitrary point inside A as the center, with a radius of r. The open ball is then the union of closed balls? Could the closed balls have a radius of r-1/n for every integer n, provided that r-1/n > 0? – john doe Sep 19 '13 at 06:40