Existence of half-planes with respect to regular open sets of the Euclidean plane

I initially asked this question at math.stackexchange.com but there was no reaction, so I thought this may be a good idea to transfer it to mathoverflow.net

Let $\langle\mathrm{r}\mathscr{O},\mathord{\subseteq}\rangle$ be the complete Boolean algebra of open domains (regular open sets, these that are equal to the interior of their closure: $a=\mathrm{int}(\mathrm{cl}(a))$) of a topological space $\langle \mathbb{R}^2,\mathscr{O}\rangle$ (i.e. the Euclidean plane with the standard topology). Let $\mathcal{C}$ be the family of all element of $\mathrm{r}\mathscr{O}$ which are convex in the standard sense, i.e. for any points $p_1,p_2\in c$, the closed interval $[p_1,p_2]$ is a subset of $c$. Let for any regular open set $\mathrm{hull}(x)$ be the convex hull of $x$, which is the smallest convex set $c\in\mathrm{r}\mathscr{O}$ such that $x\subseteq c$. We define $\mathrm{hull}$ as the special case of the closure operator on the Boolean algebra: $$\mathrm{hull}(x)=\mathrm{int}\bigcap\{c\in\mathcal{C}\mid x\subseteq c\}\,.$$

My question concerns the existence of euclidean half-planes with respect to elements of the Boolean algebra. Let $H$ be the set of all half-planes in $\mathbb{R}^2$ without $0$ and $1$ ($\emptyset$ and $\mathbb{R}^2$ itself). $H$ may be defined in the following way: $$H=\{x\mid x\in\mathcal{C}\wedge-x\in\mathcal{C}\}$$ where $-$ is the complement operation, in this case for $x$ it is $\mathrm{int}\,(\mathbb{Real}^2\setminus x)$. Let $+$ be the Boolean sum in the algbera $\mathrm{r}\mathscr{O}$: $x+y=\mathrm{int}(\mathrm{cl}(x\cup y))$, $\cdot$ the Boolean product: $x\cdot y=\mathrm{int}(x\cap y)$

It seems obvious that in case we take $a,b,c\in\mathcal{C}\setminus\{0\}$ such that $a\cdot \mathrm{hull}(b+c)\neq 0$ then there is a half-plane $h\in H$ such that for all $x\in\{a,b,c\}$: $$h\cdot x\neq 0\neq -h\cdot x\,.$$ But can we replace in the above convex sets with arbitrary regular open sets? I.e. is it true that for all non-zero $x,y,z\in\mathrm{r}\mathscr{O}$ such that $x\cdot\mathrm{hull}(y+z)\neq 0$ there is a half-plane $h$ such that both it and its complement meets each one of $x,y,z$? I will be grateful for any hints of a solution.

It seems that if you replace $x,y,z$ by their hulls and apply the `obvious' claim to them then you get the desired half-plane $h$. Namely, if $h$ or its complement contains, say, $x$ then it surely contains $\mathop{\rm hull} x$.
• I think this is not an answer since I am looking for a half-plane $h$ which meets every one of $x,y,z$, i.e. such that $h\cdot x\neq 0\neq-h\cdot x$ and $h\cdot y\neq 0\neq-h\cdot y$ and $h\cdot z\neq 0\neq-h\cdot z$. If you replace the sets with their convex hulls in the 'obvious' claim that you only now that there is $h$ which meets each one of the hulls, not the regions themselves. – Mad Hatter Oct 24 '14 at 8:37
• As I told, if a half-plane contains $x$ then it contains $\mathop{rm hull} x$, due to your definition of a hull. Thus neither the obtained half-plane nor its complement contains $x$ which is what you need. The same holds for $y$ and $z$. – Ilya Bogdanov Oct 24 '14 at 11:07
• Surely it is. If a common point $A$ of $a$ and $\mathop{\rm hull}(b+c)$ lies in $b$ then the boundary line of $h$ is a line connecting $A$ with some point of $c$; othersise $A$ lies on a line containing some points of $b$ and $c$, and this is a desired boundary line. – Ilya Bogdanov Oct 24 '14 at 14:52