6
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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 $C$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Empty Jul 3 '17 at 7:40
  • $\begingroup$ @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 $\endgroup$ – Hanul Jeon Jul 3 '17 at 8:11
  • $\begingroup$ Yes..But you have taken here $C$ as a closed set $\endgroup$ – Empty Jul 3 '17 at 8:33
  • $\begingroup$ @s717717 I've found my answer has a typo. Is my answer fine now? $\endgroup$ – Hanul Jeon Jul 3 '17 at 8:37
  • $\begingroup$ 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..? $\endgroup$ – Empty Jul 3 '17 at 19:53
1
$\begingroup$

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 $\epsilon$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – john doe Sep 19 '13 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.