20.2: Polygonal sets
 Page ID
 23711
Elementary set on the plane is a set of one of the following three types:

onepoint 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 onepoint 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 onepoint 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 onepoint 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.