5
$\begingroup$

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: $$A\leqslant B\oplus C\ \mbox{and}\ B\cdot C\neq 0\Longrightarrow A\leqslant B-C\ \mbox{or}\ A\leqslant C-B\,.$$
$\endgroup$
4
+100
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Eric, where does existence of an open ball $U$ in which $D\cup E$ is dense come from? $\endgroup$ – Mad Hatter Jul 15 '15 at 22:13
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jul 16 '15 at 6:58
  • $\begingroup$ Ah, OK - I misinterpreted <i>dense in</i> property. Thank you very, very much! $\endgroup$ – Mad Hatter Jul 17 '15 at 9:10
  • $\begingroup$ 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. $\endgroup$ – Mad Hatter Jul 23 '15 at 12:41
  • $\begingroup$ 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 $x\in C$. $\endgroup$ – Eric Wofsey Jul 23 '15 at 12:45

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.