Working on the 2D plane, I'm looking for an elegant proof of the fact that if four regions $A$, $B$, $C$, $D$ delimited by axis-parallel rectangles are such that:
- $A$ intersects $B$,
- $B$ intersects $C$,
- $C$ intersects $D$,
- $D$ intersects $A$,
- $A$ does not intersect $C$, and $B$ does not intersect $D$
(in other words, the intersection graph of these four rectangles is a cycle) then these four rectangles delimit another rectangular region of points that do not belong to any of $A$, $B$, $C$, or $D$.
It looks obvious on such a picture. But surely there must be some elegant way to prove this fact.
My current approach would be to define each region $R=([R_{x},R_{x'}], [R_{y}, R_{y'}])$ as pair of intervals (the projection on each axis). Then justify that $[A_{x},A_{x'}]\cap[C_{x},C_{x'}]\ne\emptyset \iff [A_{y},A_{y'}]\cap[C_{y},C_{y'}]=\emptyset$ (otherwise $A$ would intersect $C$ or $B$ would intersect $D$), likewise for $B$ and $D$. Then by symmetry fix an order on $A$, $B$, $C$, $D$, so it looks like in the picture with $A_{y'}< C_{y}$ and $B_{x'}
Maybe there is a Jordan-curve-like theorem I could use?