Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble showing the following.

Let $F \subset O \subset \mathbb{R}$, where $F$ is closed and $O$ is open. Prove that there is an open set $U$ such that $F \subset U$ and $\bar{U} \subset O$.

It seems so trivial, but I can't get a start on this question. Can I start with intervals?

share|improve this question
1  
Is it somehow possible to give a slightly more meaningful title to this question? –  Asaf Karagila Mar 22 '12 at 23:51
    
@AsafKaragila: How about this? –  Nate Eldredge Mar 22 '12 at 23:57

2 Answers 2

up vote 3 down vote accepted

Every open subset of $\mathbb{R}$ is a countable union of disjoint open intervals, so you can at least start by considering $F\subset O=(a,b)$ for $a<b$.

Now let $a'=\inf F$ and $b'=\sup F$. Then $a'> a$, otherwise $F$ would not contain one of its accumulation points, and similarly $b'<b$.

This should be enough to help you find your new open set $\bar{U}$.

(It's maybe not immediate to go from this case to the general case, but I don't think it's too hard either)

share|improve this answer

Since $O$ is open, for every $x \in F \subset O$ there exists $\varepsilon(x) > 0$ such that the ball $B_{\varepsilon(x)}(x)$ is contained in $O$. Then $U := \bigcup_{x \in F} B_{\varepsilon(x)/2}(x)$ satisfies $F \subset U \subset \overline U \subset O$.

share|improve this answer
    
Does this work in every metric space? –  you Mar 23 '12 at 0:04
    
Good question. Now I am not even sure whether it works in $\mathbb R$ because the closure of the union of open balls is not obviously equal to the union of the closed balls of the same radius... –  marlu Mar 23 '12 at 1:00
    
I think that should read "not always equal": take the sequence of balls $(-1+\frac{1}{n},1-\frac{1}{n})$. The union of their closures is $(-1,1)$, but the closure of their union is $[-1,1]$. I also tried coming up with a counter example to your claim that "$\bar{U}$ is a strict subset of $O$" without success, but I haven't been able to prove it either... –  you Mar 23 '12 at 2:12
    
@marlu: If $F$ is compact you can extract a finite open cover from your $U$ and then argue that the closure of the union of a finite number of open balls is the union of the closure. –  Louis Mar 23 '12 at 11:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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