If you understood the proof, try proving a similar statement for "letters T"

The solution is actually incredibly intuitive, very nice

How to translate every point of 1*1 square to 0..1 interval - it's super easy. If (x,y) position of a point on the square, then we can translate it to L=(x+y)/2, where L will be position on a line. We will have problems with points like (a,b) and (b, a) that are translated to the same point on a line, but they can be solved with modified one, like L = x*0.4 + (1-y)*0.4

Surprisingly straightforward proof, I guess because you defined the figure-eight shapes to have nearly the exact properties needed for the proof. (Other than that the two halves don't have to be circles, and that they can touch each other by more than a single point).

@MetaMaths

Жыл бұрын

I had a similar problem about "letters T" - there is also at most countable infinity of them. But eights make a better video I thought.

@MrCheeze

Жыл бұрын

@@MetaMaths Oh yeah, that seems less straightforward... seems like it should be true, but with more fiddly details to make sure that the two "regions" of each T can't overlap at the same time. I agree this is the better video, although both are interesting.

My solution: We can partition the set of eights into a sequence of sets S indexed by the integers. We let S_n be the set of all eights we have drawn whose smallest radius is between 1 / 2ⁿ and 2 / 2ⁿ (inclusive). All these must be at least 2 / 2ⁿ long and can thus not fall inside the smaller circle of another one. We then introduce a grid of squares of side lengths 1/2ⁿ. Each smaller circle will contain at least one of the grid points. This gives us a way to index all eights in S_n by a pair of integer, which means it is countable.

1:17 no we don’t, but we can

Draw an 8, where 2 sircles are the same and they are tangent at the origin. inside the 2 balls draw another 2 8s. continue utntil the end of time. each 8 can be described as combination of l and r, or 1 and 0. for example, the first 8 is just nothing. the 1 we drew in the left circle is 1, the one in the left circle of it is 11, but the one in the right circle of 1 is 10. let's add to each name of the 8s a 2, so that all the names are the same length. the first 8 is 22222222... the one on the left is 12222222222.... and so on. every combination of 0, 1 and 2 is used. there are uncountably ,amy of theese, therefore there can be uncountably many 8s.

@MetaMaths

Жыл бұрын

The set of all finite binary sequences is countable

@Detteraleon

Жыл бұрын

Fair

@MetaMaths

Жыл бұрын

@@Detteraleon Did you understand the proof discussed in the video ? Or some part is unclear ?

@Detteraleon

Жыл бұрын

@@MetaMaths i understood it as well as i could at 2 am

Why is six actually not that afraid of seven? Because seven 8s nine a countably infinite number of times

Basically the proof amounts to QxQ -> Q

@MetaMaths

Жыл бұрын

everything is hard before it is simple

@lloydgush

Жыл бұрын

@@MetaMaths Yes. I was pondering on how to label them unequivocally. Fortunately axiom of choice can be used, lol! "They aren't congruent, so they encompass different points, just pick a different set of points for each, one from each circle, OK?"

@MetaMaths

Жыл бұрын

@@lloydgush well, yes, we need to pick two rational points for each figure (one point for each circle ), and it is "by observation" we understand that no two such points can belong to two different figures simultaneously

@lloydgush

Жыл бұрын

@@MetaMaths Well, they aren't congruent, if they were they would be the same tangential circles. Not to mention they can't intersect, so had to be congruent when you have to be completely incongruent. But using the axiom of choice feels like a cheat, lol! It feels like we should have a method for picking the unique pairs of point in a regular fashion, but I can't think of any.

@JadeVanadiumResearch

Жыл бұрын

@@lloydgushActually AOC isn't needed here, because we're selecting representatives from Q^4, which is countable. Countable sets are well orderable because they are in bijection with the natural numbers, and the natural numbers are well ordered basically by definition. The min function of a well order induces the required choice function, which is constructive (and hence doesn't need Choice). That being said, AoC is a totally valid way to induce the required choice function, I just wanted to share this interesting piece of math.

simple proof of uncountably infinitely many figure eights: tile the plane with circles along all the grid lines. Then fit circles between those circles on the grid vertices. repeat this infinitely many times, down and down, until the entire plane has been filled with circles. [Assumption: there are now uncountably infinitely many circles]. Then, inside each circle, fit a unique figure eight by choosing one real number between 0 and 1/2; and another between 0 and 2pi to decide how far in and at what angle the tangent of the figure eight will be drawn, where both of its circles are also tangent to the circle they are drawn in. There are now either uncountably infinitely many figure eights, or the assumption was wrong. Every circle can be classified precisely by its center, since the generating process is so regular. Every grid coordinate (i+sqrt(2), j+sqrt(2)) for integers i,j will have a circle (countably infinte via the diagonal argument). Every grid coordinate (i,j) will also have a circle, as will all four places you could fit circles between the first two steps, and so on, at every step (infinitely many steps) there are an exponentially growing countable infinity of circles (N^2 cell circles, N^2 grid point circles, 4N^2 fitted, and after that always 3x the previous amount). I don't have the skill to prove it, but is it really possible that an exponentially growing procedure performed infinitely often where at each step you introduce another countable infinity could ever NOT be uncountable?

@JadeVanadiumResearch

Жыл бұрын

The proof in the video shows why this doesn't work; no matter what construction you use, if all the figures are pairwise disjoint, then you can select a unique pair of rational points for each figure, which shows that your set of figure eights is only countable. This is important, because valid proofs cannot give incorrect results. If two proposed proofs give contradictory answers, then one of them must be invalid. I have two other proofs that are more specific to your construction. Firstly, a countable union of countable sets is countable. At each stage in your construction, you have a set S_n of figure eights in the plane. Because S_n is countable at each stage, there is a bijection from S_n to the set of naturals N, but we can compose this with a bijection to the set of ordered pairs (k,n). So for each integer n, we have a bijection from the set {(k,n) : k is in N} onto the set S_n. Taking the union over all n, the domain becomes the entirety of N^2, and the range becomes the union of all the S_n, proving that the union across all the S_n is at most countable. My second proof is a more explicit bijection; we'll consider just one of the N^2 many tiles, and prove that there are only countably many figures in each tile. Since the construction of each tile is basically the same, this bijects your entire constriction with N^3, which is countable. For the initial stage, we draw a figure 8 inside the given tile. For successive stages, for each 8 in the previous stage we fit a new 8 into the top circle and another one into the bottom circle. However, we can track the position of the 8 with a binary string: if we go to the top circle then we use a 1, and if we go to the bottom circle we use a 0. Each stage tacks on this extra bit of information to the previous stage, so each 8 in each stage is uniquely assigned a finite binary string. If we interpret these strings as integers written in base 2, then the entire structure is proven to be countable. The high level intuition is that an uncountable set occurs when you need an infinite amount of information to specify what each object actually is. Real numbers are uncountable because we need an infinite number of binary bits to completely specify what the number is. The integers (and your figure 8s) only need finitely many bits to specify what they are, so they are countable. Although the 8s further along in your construction need progressively more bits to define their position, it's also true that very large integers need progressively more bits to define what they are, so this isn't enough to make the completed set uncountable. The ultimate reason this construction doesn't work is that none of the 8s, individually, require an infinite amount of information to define it.

wait I am with cantor, how the hell we have R -> R×R ??

@MetaMaths

Жыл бұрын

consider decimal expansions )

@gustavosilveirafrehse1508

Жыл бұрын

@@MetaMaths hmmm, I think i got it. For a number Z, can we get the even index digits of Z to be X and the odd index digits to be Y? then f(Z) = (X, Y) and (I think) f is bijective.

@MetaMaths

Жыл бұрын

@@gustavosilveirafrehse1508 you are as smart as Cantor, sir )

@gustavosilveirafrehse1508

Жыл бұрын

@@MetaMaths hahah that's so cool!! thanks for the follow up!