When looking up the definition of polygon, Wikipedia tells me:

In elementary geometry, a polygon /ˈpɒlɪɡɒn/ is a plane figure that is bounded by a finite chain of straight line segments closing in a loop to form a closed chain or circuit.

Does this definition include sets of vertices like $\{(0,0),(5,0),(6,0),(0,0)\}$, which can be displayed in just one dimension?

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    $\begingroup$ That is not a polygon in conventional terminology, although some authors may label such vertex sequences as degenerate polygons because it is a limiting case of polygons being flattened out. $\endgroup$
    – hardmath
    Commented May 6, 2016 at 11:33
  • $\begingroup$ Thank you, so it is not perfectly defined? Or are such authors simply mistaken? :D $\endgroup$ Commented May 6, 2016 at 11:35
  • $\begingroup$ I haven't looked at the Wikipedia article this morning, so it may give the description you cite at the top of the article and a formal definition in a separate section below. However an important idea tp include in a definition is that the loop or closed circuit formed by edges (straight line segments) does not self-intersect until the last edge ends where the first one began. $\endgroup$
    – hardmath
    Commented May 6, 2016 at 11:51
  • $\begingroup$ The lack of self-intersections of the "boundary" is suggested by the phrase that says it is "a plane figure that is bounded by a ... loop". A simple closed path separates the plane into an "interior" region and an "exterior" region, and the reference here is to the bounded (finite) interior region as a polygon. $\endgroup$
    – hardmath
    Commented May 6, 2016 at 11:55
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    $\begingroup$ I would say you looked up a (or one) definition of a polygon, not the definition. As it allows what we might called "degenerate" polygons, it might not be a very good definition. $\endgroup$ Commented May 6, 2016 at 17:42

3 Answers 3


Your set of vertices satisfies all the terms of the definition, so it is technically a polygon by that definition. Some would call it a degenerate polygon.

To disallow degenerate polygons, you will need to modify the definition, adding additional constraints.

EDIT: in the original post, I claimed that adding the condition that there exist at least non-collinear segments would remove the degenerate polygons. This is false: see comments.

  • $\begingroup$ So the empty set is plane figure and it is bounded by such a chain? $\endgroup$
    – miracle173
    Commented May 6, 2016 at 11:42
  • $\begingroup$ Is it bounded by straight line segments? $\endgroup$ Commented May 6, 2016 at 11:46
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    $\begingroup$ Asking that the line segments not be collinear is too much - if you, for instance, drew the letter H on a grid, you would find that the tops of either side of the H were collinear. That is, having two line segments from one line is okay, so long as the segments are separated from each other. Exactly what you want to say depends on what you consider degenerate, but a reasonable condition would be to ask that any two line segments intersect at at most one point. (More restrictive would be to demand that any two line segments may only intersect at their endpoints) $\endgroup$ Commented May 6, 2016 at 14:55
  • $\begingroup$ @MiloBrandt by degenerate I meant no interior $\endgroup$
    – Andrea
    Commented May 6, 2016 at 15:15
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    $\begingroup$ For what it's worth, I already took "two or more straight line segments must be non-colinear" to mean "there exist two segements which aren't colinear" when I read it the first time. It's a necessary condition for the polygon to have an interior, but it might not be sufficient to define a "proper polygon", that it allows for a polygon with a horrible mess at one end :-) $\endgroup$ Commented May 6, 2016 at 15:42

These things are not universally defined. In some contexts it would make sense to admit your example as a polygon, and in others it would not.

An example of the first context would be a discussion of a computer algorithm for detecting whether a point was interior to the polygon, or for calculating the area or the convex hull of a polygon. One would expect the algorithm to work reasonably even for a degenerate polygon.

An example of the second context would be the study of plane tilings or tessellations, where degenerate polygons are uninteresting as tiles, or a discussion of the triangulation of manifolds into simplices, where the triangles are expressly required to be non-degenerate.

Typically (but not always) each author will state the particular definition or at least make a remark like “we exclude degenerate polygons”.

  • $\begingroup$ Perfect lesson for Robert Hönig. $\endgroup$
    – Piquito
    Commented May 6, 2016 at 11:55

Seems to fit that definition.

However facets, being parts of polytopes such as polygons and polyhedra, shouldn't generally line up in the dimension below that of there dimension. For example edges shouldn't be collinear, faces shouldn't be coplanar, and points shouldn't be in the same place as each other. Note that star polygons are still fine.

It might be desirable to be able to switch between an "abstract polytope" like approach where polytopes are made of a set of such parts, for example a square has 4 edges, 4 vertices, and a face. And an approach where the polygon is made of the points that form its edge, a little like how a circle is defined. This causes ambiguity if adjacent edges are aloud to be in a line, or if points can be the same place as each other.

On a somewhat related note, tilings generally avoid 2 edge vertices.


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