20.2: Polygonal sets

Elementary set on the plane is a set of one of the following three types:

• one-point set;
• segment;
• solid triangle.

A set in the plane is called polygonal if it can be presented as a union of a finite collection of elementary sets.

Note that according to this definition, the empty set $\emptyset$ is a polygonal set. Indeed, $\emptyset$ is a union of an empty collection of elementary sets.

A polygonal set is called degenerate if it can be presented as union of finite number of one-point sets and segments.

If $X$ and $Y$ lie on opposite sides of the line $(AB)$, then the union $\blacktriangle AXB\cup \blacktriangle BYA$ is a polygonal set which is called solid quadrangle $AXBY$ and denoted by $\blacksquare AXBY$. In particular, we can talk about solid parallelograms, rectangles, and squares.

Typically a polygonal set admits many presentations as a union of a finite collection of elementary sets. For example, if $\square AXBY$ is a parallelogram, then

$\blacksquare AXBY=\blacktriangle AXB\cup \blacktriangle AYB=\blacktriangle XAY\cup \blacktriangle XBY.$

Exercise $\PageIndex{1}$

Show that a solid square is not degenerate.

Hint

Assume the contrary: that is, a solid square $\mathcal{Q}$ can be presented as a union of a finite collection of segments $[A_1B_1], \dots, [A_nB_n]$ and one-point sets $\{C_1\}, \dots, \{C_k\}$.

Note that $\mathcal{Q}$ contains an infinite number of mutually nonparallel segments. Therefore, we can choose a segment $[PQ]$ in $\mathcal{Q}$ that is not parallel to any of the segments $[A_1B_1], \dots, [A_nB_n]$.

It follows that $[PQ]$ has at most one common point with each of the sets $[A_iB_i]$ and $\{C_i\}$. Since $[PQ]$ contains infinite number of points, we arrive at a contradiction.

Exercise $\PageIndex{2}$

Show that a circle is not a polygonal set.

Hint

First note that among elementary sets only one-point sets can be subsets of the a circle. It remains to note that any circle contains an infinite number of points.

Claim $\PageIndex{1}$

For any two polygonal sets $\mathcal{P}$ and $\mathcal{Q}$, the union $\mathcal{P}\cup\mathcal{Q}$ as well as the intersection $\mathcal{P} \cap \mathcal{Q}$ are also polygonal sets.

Proof

Let us present $\mathcal{P}$ and $\mathcal{Q}$ as a union of finite collection of elementary sets $\mathcal{P}_1,\dots,\mathcal{P}_k$ and $\mathcal{Q}_1,\dots,\mathcal{Q}_n$ respectively.

Note that

$\mathcal{P}\cup\mathcal{Q} = \mathcal{P}_1 \cup \dots \cup \mathcal{P}_k \cup \mathcal{Q}_1 \cup \dots \cup \mathcal{Q}_n.$

Therefore, $\mathcal{P}\cup\mathcal{Q}$ is polygonal.

Note that $\mathcal{P}\cap \mathcal{Q}$ is the union of sets $\mathcal{P}_i\cap \mathcal{Q}_j$ for all $i$ and $j$. Therefore, in order to show that $\mathcal{P}\cap \mathcal{Q}$ is polygonal, it is sufficient to show that each $\mathcal{P}_i\cap \mathcal{Q}_j$ is polygonal for any pair $i$, $j$.

The diagram should suggest an idea for the proof of the latter statement in case if $\mathcal{P}_i$ and $\mathcal{Q}_j$ are solid triangles. The other cases are simpler; a formal proof can be built on Exercise 20.1.1.

A class of sets that is closed with respect to union and intersection is called a ring of sets. The claim above, therefore, states that polygonal sets in the plane form a ring of sets.