Let $\Omega$ be an open connected set in $\mathbb{C}$, not necessarily bounded. Does there exist a countable locally finite cover of $\Omega$ consisting of only open discs $\{ B(z_i, r_i): i\geq 1\}$ satisfying the following properties:
$(i)$ $\overline{B(z_i, r_i)}\subset \Omega$.
$(ii)$ $B(z_i, r_i) \cap B(z_{i+1}, r_{i+1})\neq \emptyset$ for all $i\geq 1$?
This statement is used in a theorem that states that the second de Rham cohomology group of such an $\Omega$ is $\{0\}$.
It will be great (sufficient as well) if anyone can give me a proof (or source) of the latter statement in terms of the differential forms involving only $\mathbb{R}^n$.
Thank you.
Suppose $\{ B_n: n \in \mathbb{N} \}$ is a locally finite covering of $\Omega$ by open connected sets (they needn't be discs) satisfying conditions (i) and (ii). Let $K$ be a Jordan curve lying within $\Omega$ and let $C$ and $D$ be the two parts of $\Omega \setminus K$ lying inside and outside of $K$. $K$ is compact so there is an $N$ such that $\bigcup_{n > N} B_n$ is disjoint from $K$. This union is connected (by condition (ii)) and so must lie within one of $C$ or $D$. If $D$ then $C$ is covered by $B_1, ...,B_N$ and, by (i), $\overline{C} \subset \Omega$, and vice versa. So one of $C$ or $D$ has no boundary points except in K, i.e. must be a whole component of $\mathbb{R}^2 \setminus K$. It follows that $\Omega$ is either the whole plane with at most one hole or is simply-connected. If the $B_n$s are bounded then it can only be simply connected.
If the $B_n$ are required to be disks then there is no such covering for an annulus even if condition (i) is dropped. Taking $K$ to be a concentric circle, one of the two components of the rest of the annulus would have to be covered by finitely many disks lying within the whole annulus. Given N such disks it will always be possible to find a concentric circle within the annulus but near enough to the boundary that less than $1/N$ of its circumference can be in any disk, and so it is not covered.