Building with infinite polygons

[@potaatobaked7013]

Alright, here's my reasoning as to why uncountable collections are not infinitely-buildable: 1. an uncountable set cannot be mapped to a countable set 2. For any infinitely-buildable collection, we can map each point to the step it was first added 3. There are always a countable number of steps since they are integers and the integers are countable 4. Since an uncountable set cannot be mapped to a countable set, and we can map all infinetly-buildable collections to a countable set, all infinitely-buildable collections must be countable

[@TheGrayCuber]

This is great! There is one extra step we should take though: we want the mapping to be one-to-one, but if two points were introduced in the same step then they’ll be mapped to the same integer. But this is a short fix using the fact that there are only countably many ordered pairs of integers

[@ijpthegreat]

Wow, i just realized now that youre the really big blind solver, this is such a cool channel pivot, love this math/geometry stuff

[@rubenvanderark4960]

I love this idea, it's nice to see you decided to make a second part going more in depth on it!

[@felixstrider]

i take issue with the 1-point build, its like saying the sine of infinity is 0 because thats where it starts

[@TheGrayCuber]

That build does not start with 1 point, nor does it have just the 1 point at any finite step. It results from that point being the only point with a coefficient that doesn't converge to 0

[@milojacquet7507]

It seems like it would be possible to generalize this construction to uncountably-buildable sets by allowing uncountably many steps. Since the convergence of a point does not depend on the order of the steps, just the cardinality of the sets of positive and negative steps the point occurs in, this should not cause problems. Using this definition, I believe every set of points where each point has at most countable multiplicity (positive or negative) is uncountably-buildable (which I will use here in a sense that includes finitely- and countably-buildable). By partitioning the set into rational subcollections as in the previous video, you can use the technique from this video to build each one and take the union of their sets of steps, which cannot overlap on any points. This result cannot be extended to the case where points can have uncountable multiplicities. For instance, the point set that contains just 1 with uncountable multiplicity is not uncountably-buildable since by the pigeonhole principle, at least one other point a rational angle away from 1 would also need uncountable multiplicity.

[@okboy4921]

I dont think the reasoning at 6:42 works generally, because we could have more points at every step so it would not converge. For example for every point expect first make the negative version of it and 2 new points this way number points is strictly increasing so it cant converge to the first point.

[@TheGrayCuber]

I think that just depends on how convergence of the build is defined. If you require convergence of the number of points, then sure it doesn’t work. But if you just use convergence of each point itself, then it does. I used the second definition since the results are more interesting.

[@gavinkarnatz9842]

10:33 and maybe an 11-gon (shows 12-gon)

[@TheGrayCuber]

shhhhh my software didn’t have an 11-gon option and I didn’t want to manually make one

[@NStripleseven]

10:31 I don’t think there’s any need to check whether you’d hit a point you’ve already used in this way. You could probably just get away with using the next prime every step.

[@TheGrayCuber]

That may work in some cases, but it can cause problems in others. Let p(n) be the nth prime, and t(n) be the nth triangle number. Set each a_i to be 1/(p(t(i-1)+1)) of the circumfernce. Always picking the next prime with this set of points will result in a conditional point at 0 of the circumference.

[@anthonycannet1305]

I don’t think I understand how it’s possible that an infinite sequence of steps that don’t change the average of all the points can result in something that has a different average. If I have an average score of 0 and (don’t) turn in infinite tests that each scored 0 points, how does my grade go up? I just don’t get how you can construct the single point even after infinitely many steps. If there are infinitely many steps than you’ve never actually completed the construction, and if we assume you’re at the end of the infinite construction after a series of infinite steps that all increased the number of points while maintaining a zero average, then there are infinitely many points spread around the circle. Just because one of those points was there the whole time doesn’t mean any of the following points don’t exist. Also why can’t you just define the empty collection of points as being centered? Logically speaking if there are no points on the circle then the average is at the center. I know mathematically there are divide by zero issues when calculating the average of nothing but if a finitely buildable construction is centered then adding and removing a polygon as a 2 step construction must be centered…

[@rarebeeph1783]

This is a complication that comes up any time you want to do an infinite number of steps of some process. With normal inductive reasoning, you can say, "It's centered if you do one step, and whenever it's centered, it'll remain centered after one more step, so it'll be centered after any finite number of steps." But with infinite sequences, you need an additional base case "at infinity", because induction can never reach infinity. Consider intersections of open intervals. For any open interval (a,b), its intersection with another open interval, will also be an open interval. But if you take the intersection of the infinite collection {(-1/n, 1/n) for all natural numbers n}, you'll find that the only point in all of them is 0, and so the total intersection is just the singleton set containing 0--distinctly no longer an open interval!

[@rarebeeph1783]

To more directly answer your question: the way the infinite constructions are defined was carefully chosen such that it "looks finite up close". That is, if you zoom in on any point, you will only ever see that point being affected finitely many times. So although naively you would never actually complete the construction, you can pick any point on the entire circle and see that point completed after only finitely many steps. What the result then ends up being, is the collection of whatever coefficients the individual points settle at. Under the given construction, we see infinitely many points added and infinitely many points removed, yes, but since every point but the one is either not affected at all, or added exactly once and removed exactly once, we have that the final collection is just the single point; everything else cancels.

[@TheGrayCuber]

I do consider the empty set to be centered. Yes, there is divide by 0 problems, but I usually think of centeredness as summing to the center, not averaging to the center. I just decided to use averaging for the video since it is a little more straightforward and the empty case was uninteresting enough to leave out.

[@lukatolstov5598]

1:16 You mean digon.

[@danielrhouck]

Well y₀ for the final answer seems to only be because the process as defined is only countable. But that does not seem necessary. Suppose I say that for every time t, in [0, 1), I will add a digon at angle πt. This builds the whole uncountable circle. We need to update our definitions of convergence, though, to account for multiple infinities. It seems easy enough if we just want to change it into positive and negative cardinalities for the weights, disallowing when a point appears in the same cardinality of positive and negative steps. But if we want something more complicated like measure theory, I don’t know enough to design definitions for it.