dimension-theory-analysispolygons
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?
We can make use of the symmetry here and use Pythogoras theorm to Solve:
Observe that:
QR = PO = 22 {Opposite sides of a rectangle}
PT + TO = 22
TO = 22- PT = 22 - 10 = 12
Similarly,
PQ = OR = 13 {Opposite sides of a rectangle}
OS + SR = 13
OS = 13 - SR = 13 - 8 = 5
Now In right Traingle TOS, rt angled at O,
$TO^2 + OS^2 = ST^2 {Using Pythogoras Theorm}$
$12^2 + 5^2 = TS^2$
$TS^2 = 144 + 25 = 169$
Hence, TS = 13
Now you can find the perimeter.
Instead of the definition on wiki, a more authoritative reference is Hilbert's The foundations of geometry. In $\S 4$, a polygon is defined as follows
A system of segments $AB, BC, CD, \ldots, KL$ is called a broken line joining $A$ with $L$ ...... The points lying within the segments $AB, BC, CD, \ldots, KL$, as also the points $A,B,C,D,\ldots,K,L$, are called the points of the broken line. In particular, if the point $A$ coincides with $L$, the broken line is called a polygon...
The key part is the starting point $A$ coincides with the ending point $L$. This is the same as the requirement of closed polygon chain on wiki.
By this definition, a single line segment cannot be a chain (unless it reduces to a single point).
A polygon isn't just a collection of its points. In addition to its points, it also contain the geometrical/combinatorial information like where are the vertices and how are the vertices connected.
Best Answer
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.