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
\(\blacksquare AXBY=\blacktriangle AXB\cup \blacktriangle AYB=\blacktriangle XAY\cup \blacktriangle XBY.\)
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.
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.
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.