# Is the set of all unique (convex) polygons countable? If so, by what bijection to the natural numbers?

Polygons are, in this question, defined as non-unique if they similar to another (by rotation, reflection, translation, or scaling).

Would this answer be any different if similar but non-identical polygons were allowed? And if only if rotated/translated by rational coefficients?

Would this answer be any different if we constrained the length and internal angles of all polygons to rational numbers?

Assume the number of sides is finite but unbounded, and greater than two.

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## 1 Answer

There are uncountably many, because for example one can have rectangles with arbitrary side ratios. For your second question, if everything is constrained to be rational, there will be countably many, because a polygon is uniquely determined by its ordered collection of sides and angles.

This is part of a general fact: if $A$ is a countable set, then the collection of ordered $n$-tuples of elements of $A$ for all $n$ is still countable.

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This popped up as I was thinking of writing it. +1 :) –  Cam Jul 23 '10 at 2:14
Assuming finitely-many sides; otherwise, the collection of side-lengths becomes the (ordered) power-set of the rationals. –  BlueRaja - Danny Pflughoeft Jul 23 '10 at 2:25
Does this general fact still apply if your elements are ordered infinite-tuple elements? Because the number of sides of polygons are unbounded. Would it still be countable? –  Justin L. Jul 23 '10 at 2:37
Justin: No, because the set of all infinite sequences over $Q$ is of the same cardinality as $R$. In fact, the set of all sequences over $\{0,1\}$ is of the same cardinality as $R$---this is by the binary encoding. If you are allowing polygons with infinitely many sides (finite but unbounded is OK), then I'll edit my answer - could you please clarify? –  Akhil Mathew Jul 23 '10 at 2:40
I mean finite, but unbounded. –  Justin L. Jul 23 '10 at 6:40
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