3
$\begingroup$

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

  • 0
    Ana, to get $\LaTeX$ to render you need to get your code inside `$` signs.2012-09-04

1 Answers 1

7

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 $F$.

  • 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