1

Prove: Show that for every local compact space X holds the following:

A $\subseteq$ X is closed $\iff$ $A \cap K$ is compact, for all compact sets K.

I use the following definition of local compactness:

Definition: A hausdorff space is local compact, if every $x \in X$ has a neighborhood $U\subseteq X$ which lies in a compact set.

Marc
  • 2,074

2 Answers2

4

In a Hausdorff space, every compact set is closed (How to prove that a compact set in a Hausdorff topological space is closed?).

This establishes the direction "$\Rightarrow$", because $A\cap K$ is then closed as an intersection of closed sets.

In the other direction, let $x \in \overline{A}$ be arbitrary. Then there is a neighbourhood $U \subset X$ of $x$ such that $\overline{U}$ is compact (because $U \subset K$ for some compact (hence closed) set $K$, so that $\overline{U} \subset K$ is compact as a closed subset of a compact set (Closed subsets of compact sets are compact).

But then $A \cap \overline{U}$ is closed and $x \in \overline{A \cap \overline{U}}$, because for each neighbourhood $V$ of $x$, we have

$$ A \cap \overline{U} \cap V \supset A \cap (U \cap V) \neq \emptyset, $$

because $U\cap V$ is a neighbourhood of $x$ and $x \in \overline{A}$.

Thus, $x \in \overline{A \cap \overline{U}} = A \cap \overline{U} \subset A$, i.e. $\overline{A} \subset A$, so that $A$ is closed.

Community
  • 1
PhoemueX
  • 35,087
  • One question to the first direction: A $\cap$ K is closed, but how u know that it is also compact? – Marc Jul 03 '14 at 18:50
  • Because closed subsets of compact sets are compact. http://math.stackexchange.com/questions/716245/closed-subsets-of-compact-sets-are-compact – PhoemueX Jul 03 '14 at 19:00
  • Ah, ok thanks. And $A\cap K$ is a subset of K, which is compact. – Marc Jul 03 '14 at 19:02
2

One implication is clear. For the other note that for $x ∈ \overline{A} \setminus A$ and $N$ neighborhood of $x$ we have $x ∈ \overline{(A ∩ N)} \setminus (A ∩ N)$ in $N$.

Of course we use the facts that closed subspace of a compact space is compact and that compact subspace is closed in a Hausdorff space.

user87690
  • 9,133
  • 1
    could you explain me how you can conclude the last thing? – Marc Jul 09 '14 at 23:49
  • @Xtk: Do you mean the fact that compact subspace is closed in a Hausdorff space? If you have a compact $K$ and a point $x$ in Hausdorff space $X$ such that $x ∉ K$ then by Hausdorffness there are disjoint open $U$, $V$ such that $K ⊆ U$, $x ∈ V$. This is because a compact subset beahaves like a point. $x ∉ \overline{K}$ follows. – user87690 Jul 10 '14 at 09:12
  • 1
    I know that. But i didnt understand how u can get: $x ∈ \overline{(A ∩ N)} \setminus (A ∩ N)$. Would be nice, if u can explain it :) – Marc Jul 10 '14 at 14:57
  • 1
    Since $x ∉ A$ we have $x ∉ A ∩ N$. On the other hand it is straightforward exercise using the definition of closure showing that $x ∈ \overline{A} \implies x ∈ \overline{A ∩ N}$ for any $N$ neighborhood of $x$. – user87690 Jul 10 '14 at 15:20