#What is the cardinal of all polygons in R^2?

139 messages · Page 1 of 1 (latest)

fading lance
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I believe it would be |R|?

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fading lance
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MODULO ISOMETRY, forgot to specify

vast python
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use the fact that R2 has countable neighborhood bases

fading lance
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You mean that for any point there is a countable neighborhood basis?

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My first thought would be to tackle the problem without talking about isometries first:

Well choose a natural numar n >= 3. For n=3 there we would have to think of 3 points in the plane choosen. Each of the 3 points have 2 coordonates so we have to choose 6 real numbers ==> All triangles have cardinal R^2 x R^2 x R^2 so R^6 which is the same as R from Cantor-Bernstein-Schroeder.
The same argument can be analoguosely for any polygon with n vertices.
So for each natural >= 3 there are each R polygons. So the total number of polygons would be the cardinal of the countable union of R's

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This is my train of thought

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But the original question asks if this is the case even when we allow isometries

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In that case I am not sure

fading lance
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I don't know how topology would make its way into this problem

vast python
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well, you formulated a topological problem

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it's only natural a solution would be topological

fading lance
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Ain't topology about open sets?

vast python
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indeed

fading lance
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The problem is about cardinals and polygons

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I don't seem to see

vast python
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aren't we in R2 all the time?

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set of all (convex) polygons in R2

fading lance
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R^6 represents the cardinal of all possible triangles

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R^8 represents the cardinal of all possible quadrilaterals

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R^10 represents the cardinal of all possible pentagons

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etc.

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The cardinal of all polygons ought to be the cardinal of all these

vast python
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and why do we get cardinality of all triangles by |R6|?

fading lance
fading lance
vast python
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minor detail, not really impactful whether it's convex or all

fading lance
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Yes I believe so too

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But is there a problem with my thinking above?

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Cuz you seem confused about it still

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Well anyway the thinking I gave is only for a simpler version of the original question so it still doens't answer it fully

vast python
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if so then that's fine

fading lance
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No, for an n-gon that means there are 2*n numbers of real numbers to decide to define it uniquely

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Since an n-gon has n vertices and each vertex has 2 coordonates (x,y)

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so the number of triangles is the same cardinal as R^6

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and similar for any other n-gon

vast python
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yup you have the idea

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but now you can use that R2 is separable

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you can make this precise with injections in both directions, but the only way I know how to is topologically

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I'll assume we are counting convex polygons

  1. R2 has countable neighborhood bases (e.g the set of all disks with rational coordinates as center and rational radius)
  2. We get injection from open subsets of R2 to 2^omega - an open set is mapped to a set of basis elements that it contains
  3. From considering complements we get injection from the set of closed sets to omega (2omega to be precise)

Conversely, you can just consider triangles, as you said and so there's an injection from 2^omega to the set of convex polygons

sacred fractal
vast python
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so since there are injections in both directions, the set of all convex polygons has cardinlity |R|

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it's a "traditional" topological argument for counting geometrical shapes

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bound to run into it

fading lance
# sacred fractal what's this btw

We don't consider polygons that differ only by a translation/rotation/reflections and also scaling (which I forgot to say at the beginning... >_<) to be different. We consider them to be the same

sacred fractal
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okay

fading lance
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The original question is basically what is the cardinal of all polygons/~ , where ~ means that Polygon A ~ Polygon B <==> A is similar to B

fading lance
vast python
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countable cardinal

fading lance
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Ok...

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Give me some time to digest your proof

sacred fractal
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my argument : assume for n-gons it is |R|. Every n+1 gon can be treated as {n-gon} + some other point, fix said n-gon and place said n+1-th point anywhere on the plane. This tells us that the set of n-gons having cardinality |R| implies the same for n+1 gons, hence the set of all such polygons has cardinality |R|^|N|

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Given that the set of all line segments has cardinality |R|

vast python
fading lance
sacred fractal
vast python
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it suffices to count the convex ones

sacred fractal
fading lance
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Does it?

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Idk

sacred fractal
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aL does it?

fading lance
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That's what I am asking xd

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I am rusty on my cardinal arithmetic

vast python
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an arbitrary ngon can be cut into finitely many convex ones

sacred fractal
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right

fading lance
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Ok

sacred fractal
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is it easier to count convex gons?

vast python
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convex ones are easier for me to imagine

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if they're not convex you might run into some weirdness

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which eventually doesnt matter

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but stick with the convex ones

sacred fractal
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you overcount maybe

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but thankfully this works out

fading lance
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It's fine since they are infintie cardinals

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Overcounting ain't that scary here

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I think

sacred fractal
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idk man I got my share of ptsd...

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ill backread

vast python
sacred fractal
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we can show |R|=<|n-gons|=<|R^{2n}| too

sacred fractal
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this doesn't account for any isometries

fading lance
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also don't forget about scaling too

sacred fractal
fading lance
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Forgot to mention it previously >_<

sacred fractal
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2n not n

vast python
fading lance
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It's not

sacred fractal
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ill search it up

vast python
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oh no wait, you're right

fading lance
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I did already

vast python
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N^R would be bigger

sacred fractal
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because 2^|N|

fading lance
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c = 2^omega so c^omega = 2^(omega^2) = 2^omega = c

vast python
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R^N is the cardinality of real valued sequences

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hence |R|

sacred fractal
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it is

vast python
sacred fractal
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what about treating polygons as points in C and then defining it's polynomial as having roots at its vertices

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how does the isometries carry onto the polynomial

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horrible idea

fading lance
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Nah it ain't that bad I think

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oh yeah nvm

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even translation looks horrible

vast python
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well you dont worry about it since you quotiented with respect to congruence from the beginning

fading lance
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Yeah

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The only think I have in head so to just do every case of n=3, 4,5,6, 7... to see individually their cardinal of polygons factorized to this equivalence relation of isometries + scaling and also check the union of them

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I still am not sure even your help guys

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I also don't like the topological proof you gave aL

vast python
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that's fine 😄

fading lance
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So I'll keep thinking of a more brute force proof since I prefer them rather than the one above

vast python
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think it through and you'll see that it's correct

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funnily enough, I'd rather do topological proof because it's rigorous enough 😂

fading lance
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In your proof, when you say open set do you mean the interior of the polygon defining it?

sacred fractal
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@fading lance do you not get the cardinal arithmetic

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if you don't why don't you get it?

fading lance
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I do

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Ah

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Yeah I forgot about your proof

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But that proof is about the version of the problem without isometries + scaling

sacred fractal
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I was just going to tell you that cardinal arithmetic is backed up by bijections

sacred fractal
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|{n-gons with constraints}| >= |?|

fading lance
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What if the set of all triangles constrained are countable?

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Your induction step would work if this were teh case

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Also the induction step wouldn't work either way since you can't choose just any point as your n+1-th point to not be redundant

sacred fractal
fading lance
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Proof?

sacred fractal
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using your isometries, fix some base for your triangle

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let's say (0,0) to (0,1)

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ah wait

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look at the angles

fading lance
vast python