simply connected open as a union of convex open subsets Let $U$ be a simply connected and connected open subset of $\mathbb R^2$. Is it always possible to write
$U$ as a finite (non-disjoint) union of convex open subsets of $\mathbb R^2$?
 A: No.
Here is an example. Define
$$U = \mathbb R \times (0,2) \setminus \mathbb Z \times [1,2) .$$
This set is contractible. In fact, $\mathbb R \times \{\frac{1}{2}\}$ is a strong deformation retract of $U$.
Now assume that $U = \bigcup_{i=1}^m C_i$ with convex $C_i$ (it is irrelevant whether they are open or not). One of the $C_i$ must contain more than one of the points $z_n = (n - \frac{1}{2},\frac{3}{2}) \in U$, say $z_{n_1}, z_{n_2} \in C_j$ with $n_1 < n_2$. But then $C_j$ must contain the line segment $S$ connecting $z_{n_1}$ an $z_{n_2}$. But then $(n_1,\frac{3}{2}) \in S \subset C_j \subset U$ which contradicts the definition of $U$.
A: Here are a couple of bounded examples.

*

*Let $U=\{(x,y):0<x<1, 0<y<x^2\}.$


*Let $P=\{(x,y):1<x^2+y^2<2, 0<y\}.$
The proofs are similar, here is a proof for $P$.
Let $C=\{(x,y):1=x^2+y^2, 0\le y\}.$ If $V$ is
any convex open subset of $P$ then its closure
$K$ is a compact convex subset which intersects $C$
in at most one point. If $V_1,...,V_n$ are
finitely many convex open subset of $P$, then
their closures $K_1,...,K_n$ intersect $C$ in at most finitely many points, and hence there is some $(p,q)\in C\setminus \cup_{i=1}^n K_i.$
The set $K=\cup_{i=1}^n K_i$ is compact, hence
the distance from $(p,q)$ to $K$ is positive,
hence there are points $(x,y)$ near $(p,q)$ in
$P\setminus K$, so such $(x,y)$ are not covered
by any of the $V_1,...,V_n.$
