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Mathematics LibreTexts

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.

    截屏2021-03-02 下午4.35.33.png

    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.

    截屏2021-03-02 下午4.37.51.png

    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.


    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.


    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.


    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.

    截屏2021-03-02 下午4.40.09.png

    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.