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# 20.2: Polygonal sets

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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.