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.

Are all compact sets in $\Bbb R^n$, $G_\delta$ sets? I know that compact set is bounded and closed.

share|improve this question
    
Ana, to get $\LaTeX$ to render you need to get your code inside $ signs. –  Pedro Tamaroff Sep 4 '12 at 22:38

1 Answer 1

up vote 6 down vote accepted

In metric spaces, every closed set is $G_\delta$.

In metric spaces (more generally Hausdorff Spaces), compact subsets are closed.

Hence all compact subsets of the metric space $\mathbb{R}^n$ are $G_\delta$.


Since you already know that compact subsets of $\mathbb{R}^n$ are closed. If you want a hint on how to show closed subsets of metric spaces are $G_\delta$, move over the box below:

Let $F$ be a closed set. Define the open set $U_n = \bigcup_{x \in F} B_{\frac{1}{n}}(x)$. Show that $\bigcap_{n \in \mathbb{N}} U_n$ consist of exactly the points of $F$ and its limit points. Use the fact that $F$ is closed to conclude that this intersection is $F$.

share|improve this answer
    
My problem is that if my compact set is [2, 3], then $\bigcap_{n \in \mathbb{N}} U_n$ contains the $2-\epsilon$ too. if the $x$ in $\bigcup_{x \in F} B_{\frac{1}{n}}(x)$ doesn't go to infinity. Am I correct? –  Ana Sep 4 '12 at 23:07
    
@Ana No, it isn't in this intersection. Choose a $n$ such that $\frac{1}{n} < \epsilon$. $2 - \epsilon \notin U_n$ for this $n$. Hence $2 - \epsilon \notin \bigcap_{n \in \mathbb{N}} U_n$. –  William Sep 4 '12 at 23:10
    
is $B_{\frac{1}{n}}(x)$ an open interval lile $(2-\frac{1}{n},3+\frac{1}{n})$ ? –  Ana Sep 4 '12 at 23:15
    
@Ana Yes. So for any fixed $\epsilon$ and choosing $n$ such that $\frac{1}{n} < \epsilon$, there is no $x \in [2,3]$ such that $2 - \epsilon \in B_\frac{1}{n}(x)$. Hence $2 - \epsilon \notin U_n$, for this $n$. –  William Sep 4 '12 at 23:17

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.