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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 is a polygonal set. Indeed, 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 AXBBYA is a polygonal set which is called solid quadrangle AXBY and denoted by 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 AXBY is a parallelogram, then

AXBY=AXBAYB=XAYXBY.

Exercise 20.2.1

Show that a solid square is not degenerate.

Hint

Assume the contrary: that is, a solid square Q can be presented as a union of a finite collection of segments [A1B1],,[AnBn] and one-point sets {C1},,{Ck}.

Note that Q contains an infinite number of mutually nonparallel segments. Therefore, we can choose a segment [PQ] in Q that is not parallel to any of the segments [A1B1],,[AnBn].

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

Exercise 20.2.2

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 20.2.1

For any two polygonal sets P and Q, the union PQ as well as the intersection PQ are also polygonal sets.

Proof

Let us present P and Q as a union of finite collection of elementary sets P1,,Pk and Q1,,Qn respectively.

截屏2021-03-02 下午4.40.09.png

Note that

PQ=P1PkQ1Qn.

Therefore, PQ is polygonal.

Note that PQ is the union of sets PiQj for all i and j. Therefore, in order to show that PQ is polygonal, it is sufficient to show that each PiQj is polygonal for any pair i, j.

The diagram should suggest an idea for the proof of the latter statement in case if Pi and Qj 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.


This page titled 20.2: Polygonal sets is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.

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