Are all compact sets in $\Bbb R^n$, $G_\delta$ sets? I know that compact set is bounded and closed.
Are all compact sets in $ \Bbb R^n$, $G_\delta$ sets?
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real-analysis
general-topology
compactness
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0Ana, to get $\LaTeX$ to render you need to get your code inside `$` signs. – 2012-09-04
1 Answers
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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 $F$.
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0@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 $n$. – 2012-09-04