2
$\begingroup$

Can any of you guys think of a topological space $(X,\tau)$ and a family of subsets {As}${s \in S}$ of $X$ such that for a certain $x \in X$ you can find a subset $V$ such that $x \in V$ and {$s \in S: V \cap A_{s} \neq \emptyset$} is finite, whereas for every $W \subseteq X$, with $x \in W$, we have that {$s \in S: W \cap \mathrm{cl}(A_{s}) \neq \emptyset$} is never finite?

I thank you in advance for your replies.

  • 1
    Yeah, no, don't do that please.2013-03-02
  • 0
    If $V$ and $W$ are meant to be *open* sets, this cannot happen. The set of closures of a locally finite family is locally finite (which is not true for point-finite).2013-03-02

1 Answers 1

4

If I read it correctly you want $V$ to meet only finitely many of the $A_s$, but every set containing $x$ to meet infinitely many of the closures of the $A_s$. Without some restriction on $V$ and the $A_s$ this is easy. Just have all the $A_s$ have x as a limit point but not a member. Then $V$ is just the point $x$ and any $W$ containing $x$ will meet the closure of all the $A_s$.

  • 0
    Concrete example: the plane, $A_s = \{(t,ts): t > 0 \}$ for $s \in \mathbb{R}^{+}$ (open half lines emanating from the origin), $x = (0,0)$, $V = \{(0,0)\}$2013-03-02