# open cover

• Apr 30th 2009, 07:23 AM
manjohn12
open cover
Suppose that a metric space $X$ has a countable dense cover $Q$. Show that any open cover of $X$ has a subcover which is at most countable.

So consider an arbitrary cover of $X$ $, \{U_{\alpha}: \alpha \in A \}$. Let $\mathcal{B}$ be the set of open balls $B_{r}(q)$ where $r \in \mathbb{Q}$ and $q \in Q$. Let $E \subset \mathcal{B}$ be the set of open balls such that $U_{\alpha}$ is contained in one of the $\{U_{\alpha} \}$. From here can we say the these union of open balls is $X$ because of the denseness of $Q$? From here how do we show that the subcover is either finite or countable?
• Apr 30th 2009, 11:47 AM
albi
I'm not very good in English, but I don't think $Q$ is a cover...

I think you mean X is separable, and Q is countable and dense in X... I don't know if there is a english name for this kind of set...

First of all, you should note $\mathcal{B}$ is countable:
$|\mathcal{B}| = |Q \times \mathbb{Q}| = \aleph_0 \aleph_0 = \aleph_0$

Secondly $E$ is not a subcover of $\{U_{\alpha}: \alpha \in A \}$.

Thirdly, the idea is good, and $E$ covers X:
Let $x \in X$. Then $x \in U_\alpha$ for some $\alpha$. $U_\alpha$ is neighbourhood hence there is the ball $B_{r}(x) \subset U_\alpha$. Q is dense in X so there is $q \in Q$ such that $d(x, q) < r/3$. There exists ball $B_\epsilon(q) \in \mathcal{B}$ where $r/3< \epsilon < r/2$. Thus $x \in B_\epsilon(q)$, and $B_\epsilon(q) \subset B_r(x)$.

Of course $B_\epsilon(q) \in E$. Hence $E$ (and $\mathcal{B}$ too) is cover of X.