# (A kind of) Irreducibiliy of regular open convex sets in the Cartesian space

I am looking for a proof of the fact which is formulated at the bottom of this post. The property of regular convex sets which the fact expresses seems to be true to me, yet I have not been able to demonstrate it in a rigorous way. Any hints will by very appreciated.

Let $\mathrm{r}\mathscr{O}$ be the family of open domains (regular open sets) of a topological Cartesian space $\langle\mathbf{R}^2,\mathscr{O}\rangle$ (with the standard topology), that is: $$A\in\mathrm{r}\mathscr{O}\iff A=\mathrm{int(\mathrm{cl(A)})}.$$

It is well know that $\langle\mathrm{r}\mathscr{O},+,\cdot,-,0,1\rangle$ is a complete Boolean algebra, where: $A+B=\mathrm{int(\mathrm{cl(A\cup B)})}$, $A\cdot B=A\cap B$ and $-A=\mathrm{int}(\complement A)$ (with $\complement$ the standard set-theoretical complement operation).

In this algebra define the disjoint sum operation in the usual way: $$A\oplus B=(A-B)+(B-A)\,.$$

By a convex set $A$ I understand, standardly, a regular open set for which it is the case that for any two points $x,y\in A$: $[x,y]\subseteq A$. The fact I would like to prove is (the order $\leqslant$ is standard) the following kind of irreducibility:

Let $A,B,C\in\mathrm{r}\mathscr{O}$ be convex: $$A\leqslant B\oplus C\ \mbox{and}\ B\cdot C\neq 0\Longrightarrow A\leqslant B-C\ \mbox{or}\ A\leqslant C-B\,.$$

Let $B$ and $C$ be convex regular open sets such that $B\cdot C\neq 0$, and write $D=B-C$ and $E=C-B$. It suffices to show that the union $D\cup E$ is regular. Indeed, if it is, then $B\oplus C=D\cup E$ and $D$ and $E$ are disjoint open sets, so any connected subset of $B\oplus C$ must be contained in either $D$ or $E$.
So suppose for a contradiction that $D\cup E$ is not regular. Let $x$ be a point in $(D+E)\setminus(D\cup E)$; then there is some open ball $U$ around $x$ such that $D\cup E$ is dense in $U$. If $x\in B$, then $E$ is disjoint from a neighborhood of $x$, and so $D$ would have to be dense in a neighborhood of $x$ and hence contain $x$ by regularity. Thus $x\not\in B$, and by symmetry $x\not\in C$. By convexity of $B$, we can find a line $L$ passing through $x$ such that $B$ is contained in one of the open half-planes $V$ formed by $L$; let $W$ be the other half-plane. We thus have $D\cdot U\leq V\cdot U$, and so $E$ must be dense in and hence contain all of $W\cdot U$. If $y\in V\cdot C$, then the line from $y$ to $x$ extends into $W\cdot U$, so by convexity $C$ would have to contain $x$. Thus $V\cdot C$ must be empty. But now $B\cdot C\leq V\cdot C=0$, a contradiction.
• Eric, where does existence of an open ball $U$ in which $D\cup E$ is dense come from? Jul 15, 2015 at 22:13
• That's just the definition of $D+E$: it is the set of points which have a neighborhood in which $D\cup E$ is dense. Jul 16, 2015 at 6:58
• Eric, I have some second thoughts about the proof. Could you please explain to me the transitions in the following passage: "If $y\in V\cdot C$, then the line from $y$ to $x$ extends into $W\cdot U$, so by convexity $C$ would have to contain $x$". I am not sure if I have understood it properly. Jul 23, 2015 at 12:41
• Given any point $y$ in $V$, if you draw a line segment from $y$ to $x$, and then extend the line segment a little bit past $x$, you will enter the set $W\cdot U$. Thus $x$ lies on the line segment joining $y$ to some point in $W\cdot U$. Since $W\cdot U\leq C$ and $C$ is convex, $y\in C$ would imply $x\in C$. Jul 23, 2015 at 12:45