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

Pedro
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Ana
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1 Answers1

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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 $F$.

William
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  • 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 04 '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 04 '12 at 23:10
  • is $B_{\frac{1}{n}}(x)$ an open interval lile $(2-\frac{1}{n},3+\frac{1}{n})$ ? – Ana Sep 04 '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 04 '12 at 23:17