This one came out much longer than expected as it was a more technical video, because of that I had to omit some things I'm going to include in this comment. 1) Since the cantor set is uncountable, that means there are points in it that are NOT endpoints in any of the C_n sets (I brought this up but never acknowledged the answer). In fact, the set being uncountable means there must be irrational numbers in the set since rational numbers are countable. 2) One example of a point in the cantor set which is not an endpoint is 1/4, if you put a dot at 1/4 and move it down, it'll always be in the next C_n set but it will never be an endpoint. 3) 1/4 is known to be in the set because it has a ternary (base 3) form that does not include the number 1 (1/4 = .020202020.... in base 3). I never discussed this in the video but the cantor set consists ONLY of numbers in [0,1] that can be written in base 3 form, without the number 1. (Note: Base 2 = binary, base 3 = ternary). 4) There's a cool property about the cantor set that can be proved graphically, and if you want a challenge try to prove it. Property: For ANY number between 0 and 2 (call it p), there exists two numbers in the cantor set (call them a and b), such that a+b=p. 5) Sorry for anything that wasn't to scale, told the animator to make everything as proportional as possible but he needed room to write everything
@@l1mbo69 No time like the present. One of the neat and unique things about Math is that it ages well, no single idea ever becomes obsolete as new ideas emerge unlike with Technology, medicine, and to some extent Physics and Chemistry. That's because Math eventually became based on proof and statements of truth which remain true forever. ( if proven correctly ) That's a long way of saying you can always lean it. There is nothing uniquely special about majoring in it in college anymore. To learn what Math majors know you need 2.5 things. (1) Source material like this video, (2) Practice ( old textbooks, (maybe) brilliant.org, (maybe) myopenmath courses (if that's still around) and (2.5) patience because learning math can be hard at times in a way like a foreign language or a sport can be and requires one to push through its hard parts. In all likelihood, if your like me, you probably dont have the time/energy/resources to teach yourself all the math you wish you knew but never got to learn. But I wanted to put out there that it's not an impossible thing just in case anyone wanted to try. It's a lot easier to try something when you know its possible.
@@mattmahoney8659 thanks for the to the point structure. But I have tried, and it's not that I am terrible but I am far from good enough to justify majoring in it. I'm not going to be completely leaving it behind, though. I have decided to get a degree in Physics (and I'll try to minor in mathematics). I seem to be much better at it (judging from how I fare in international olympiads) and like it almost as much as maths. I think the issue is that Mathematics, other than logic, also involved a lot of creativity. What is it that clicks in your brain, exactly? To me it seems to be really mysterious. All the progress I have made in maths has only been by practicing similar problems till I know the tricks used. Basically I just become familiar with that particular topic and I don't feel like I am making any progress with my core mathematical intuition. Does that make sense?
@@sebastianp4023 you mean every third of the exam is removed. If every third question was removed and there were n questions, what would the last question left be?
Also, 4/9 is n the first middle third, so it’s not in their either. For people looking for a formal explanation, the set contains all number from [0,1] where the ternary (base 3) expansion has no ternary digit 1, except at the last decimal place, if it terminates. PS, sorry if you didn’t get what I said, look it up if you really wanna know.
L, after the first iteration is the segment between 0 and 1/3, inclusive. In the second iteration, LL would be the segment between 0 and 1/9, inclusive, and LR would be 2/9 to 1/3, inclusive. 1/4 falls on the segment between LL and LR which would be between 1/9 and 2/9, non-inclusive, and that section is removed in just the second iteration. 1/4 is NOT part of the Cantor set. This video has a couple mistakes, but this is not one of them. All numbers in the cantor set have a denominator that is divisible by 3, with the only two exceptions being 0 and 1. I'm open to hearing an argument that says I'm wrong and the op is right, but I'd need to see a proof. Edit: I think the fault in your logic relies on L and R being equal to 1/2 each. Both L and R are in fact representative of 1/3 each, and a silent C is the missing variable that has been removed from the set in the middle of each L and R. Edit #2: I was wrong. Read below if you want to know why. I'm editing my comment to note that I was wrong, but leaving it up as-is besides this note, in case others may have been mistaken, like me.
@@piezomofo I'm sorry to call you out, but when you said that 1/4 falls in between 1/9 and 2/9 it actually falls in between 2/9 and 3/9. 1/9 = 0.11111... 2/9 = 0.22222... 1/4 = 0.25 3/9 = 0.33333... Also if you are curious about how I found that 1/4 is in the set, it has to do with this infinite sum: 1/3 - 1/9 + 1/27 - 1/81 ... = 1/4
@@samuelthecamel no, by all means, you're right to call me out. I wasn't doing the math and was quick to throw out a reply. You're right about 1/4 being between 2/9 and 3/9, and I was wrong saying it was removed at that point. Also, I understand what you're saying about the infinite sum, but I can't wrap my head around it. I'm not saying you're wrong, in fact, it's highly likely that you're much smarter than me. I just have to imagine, though, that at some point 1/4 must be excluded because it's denominator isn't divisible by 3. I'm probably going to drive myself nuts trying to figure this out now. What you say makes sense, but it's also contrary to this idea that the only way to end up a number in the Cantor set is to be between 0 and 1 with a denominator that's divisible by 3. Either way, my mind has been sufficiently blown. I'll let you know if I make any progress, and thank you for correcting me. Now I've got a challenge.
@@samuelthecamel wow, I just read a paper from faculty.math.illinois.edu/~reznick/496-1-16-19.pdf It explained the same thing you said and went a little further, obviously easier to go further in a paper than in a UA-cam comment. The paper also says that in an informal survey (sample size 12) 9 out of 12 did not know or did not remember the fact that 1/4 is in fact in the Cantor set. That made me feel a little better about being wrong, myself. I would imagine that it stands to reason that 3/4 is also part of the set, since the set is symmetrical. I wonder what other numbers might fall within the set that are not so obvious. Anyway, thanks again for your original fun fact, and for setting me straight. I wouldn't have otherwise done the further research and may not have learned about this interesting non-intuitive point.
Yep, it is the most prominent example of such a creature. In fact, coming up with such a creature without building up from the Cantor set, or its idea, is rather difficult..
I prefer the inverse Cantor set. It has length 1, is still uncountable, and as it's length is 0+1/3+2/9+4/27... and as (1/3+2/9+4/27+..) = 1 Despite missing every number in the Cantor set, a few numbers that aren't in the inverse cantor set include 1/4 and 3/4.
@@livedandletdie i know im 10 months late, but as another comment in this comment section said, 1/4 and 3/4 are actually in the cantor set, so that is probably why the inverse cantor set does not have them
Same here. Im in high school watching math and physics videos on yt (3b1B, Mathloger, Numberphile, Veritasium...) just for fun and I understood everything in this video. So maybe I actually learned some math or it wasnt so complicated, but I cant tell :D
Fun fact: at 7:13 you mention that you could think of the elements of the Cantor set as binary numbers, with 0's in place of the L's and 1's in place of the R's. However, if you instead replace the R's with 2's, you actually get the element of the Cantor set written in base 3. That's another way to think of how to construct the Cantor set: you take all the numbers from 0-1 written in base 3, and at each step you remove all numbers with a 1 in that digit after the decimal point.
The best way to think about the length is that every C_n has 2/3 the length of C_n-1. If you start with 1 and multiply by 2/3 on every step, C_n will have length (2/3)^n. Since the cantor set is the limit of C_n as n approaches infinity, the length is the limit of (2/3)^n as n approaches infinity, wich is also know as zero.
BUT you are NOT removing 1/3 every time because you leave the end points. The "1/3" that you remove is smaller than each of the "1/3" segments that you remove.
@@lynnrathbun those end points have a Lebesgue measure of zero, they make no difference to the length. Just like the intervals (0, 1), (0, 1], [0, 1) and [0, 1] all have the same length of 1 unit. It's not 1- or 1+ or anything infinitesimaly close to 1, it's exactly 1. The length of an individual point is zero.
3:17 Fun fact: The set of all fractions with denominators that are whole powers of three (i.e. the endpoints of the Cantor set) is a countable set. This means that *almost all* the points left behind are not endpoints.
@Jorge Hernandez, all non-computable numbers are not endpoints, but there are computable non-endpoints as well. As explained in the video, every infinite sequence of L and R describes a point in the set. All endpoints will have some initial sequence and then either infinite L's or infinite R's. Any sequence with a different end behavior, such as LRLRLR... will be a non-endpoint, and plenty of those are constructable.
This is the weird part. The set seems to be just a collection of rational numbers which is countable regardless of any other caveats. So the set is countable as you are constructing it until it hits the point at infinity then it goes uncountable?
If you're constructing it by the remove-the-middle method, all the intermediate sets are uncountable. (Any segment of non-vanishing length contains an uncountable number of points.)
One takeaway from this is that "infinity" doesn't necessarily mean "all". Merely, infinity is an indicator that there is some sort of unending procedure that can be used to generate numbers, and this procedure can be used unendingly. If you remove an infinite number of numbers between 0 and 1, there will still be infinitely many numbers left over. And correct me if I'm wrong here, but the distance between negative infinity and positive infinity is the same as the distance from zero to infinity.
i dont think you're wrong depending on what you consider a distance, but honestly it's hard to tell because infinity is such a big number that it's hardly even a number at all. the distance between 0 and 1 is 1, and the distance between negative infinity and infinity is 2*infinity, and the distance between 0 and infinity is infinity/2. the problem of course is identifying whether any of these numbers are different at all after some point. of course, it would be incorrect to state that 1=infinity or infinity/2=infinity. it's like saying 0=1, or 1/2=1, even though 0/2=0, and 1=1. fractals are amazing
Mate, the end part on the fractional dimension was absolutely gold. I learnt the "box counting dimension" at university but it really glossed over my head when I was an undergrad. Your analogy is so simple to understand, that I'll probably use it to explain it to my students. Thank you :)
9:00 What baffles me is you could do the same thing with infinitely long decimal integers and discover that there are uncountably many of them. Despite the fact that there are countably many decimal integers total.
The problem then is that you'd get a real number which actually isn't an integer (or fraction if you will) thus not in that set, so you haven't reached a contradiction with this new number you made
The reason this doesn't work with the integers is that every integer has a finite length. The set of integers contains elements that are arbitrarily large, but every individual element is finite in length. If you try to apply the diagonalization method with the integers, you will end up with a number of infinite length, so it won't be a valid integer.
@@kaiblack4489 Yeah I don't know what I was thinking to be honest. Me 3 years ago not as smart as me now I guess (hopefully at least). What I described appears to be akin to 10-adic numbers, which is a representation of reals, not just integers.
It is easy to find a number written in the LRLR form. Just replace all the L's with 0's and all the R's with 2's, and you will get the number in base-3. For example, 1/3 = LRRRRRRRRRRRRRRR = 0.0222222222222222 (base-3) = 0.33333333333 (base-10) You could also go the other way around to find the LRLR form from the number. For example, 1/4 = 0.25 (base-10) = 0.02020202020202020202... (base-3) = LRLRLRLRLRLRLRLRLRLR... in the cantor set.
Why use base 3 instead of base 2? Actually you have to use base 2 instead of 3 otherwise you can't represent numbers like 4/9 which in base 3 is 0.11 or 0.102222222 both of which contains '1', but since you can't represent 1 in your method you can't represent 4/9. If you changed to base 2 it would solve that problem (i.e. L=0,R=1 like he did in the video).
@@tomerwolberg37 But that's exactly what makes it convenient. Any number which has a '1' in its base-3 representation will never be in the cantor set (4/9 for example is not in the set). It is also possible that it is an endpoint in the set (e.g. 1/3 is 0.1 in base-3, but the convenient part is that it can be written as 0.02222... which makes it a part of the set as an endpoint).
karan jan thanks for your example, yet i am a bit lost and because you clearly have a grasp of deep mathematics maybe you could help. what is this all about? Why was Cantor showing this "set"? I ask because i can somewhat follow along witht this video, but I dont know what its all about. Why are we talking about this set? What is its significance? I know its EXTREMELY significant, yet as I am ignorant i cant figure out why. Maybe you would have interest answering?
One curious thing about the Cantor set, is that, yes, it contains all the endpoints of the C0, C1, C2, ... sequence, but it does not ONLY contain those endpoints. This is clear, because there are countably many such endpoints---they can be enumerated. The Cantor set contains uncountably many points which are not endpoints of any such interval.
Bro, youre a hell of a teacher. I listen to your videos while driving (being a father and husband really limits the time frame i have to focus on myself, so i do what i can) and despite that, im able to grasp everything youre talking about clearly. I wish i had a teacher that was as good as you are, back when i was in school, maybe i would have pursued higher education.
I've just invented the non-inclusive cantor set, where instead of including the endpoints you exclude them. The full list of non-inclusive cantor numbers is as follows:
"The size of the Cantor set gets doubled when its side length is tripled." Could you please elaborate on how do you define double here? If you use cardinality, it is always that of R; if you use Lebesgue measure, it is always 0.
Having seen the video by 3blue1brown he references, I think that the size he is talking about is defined as the limit as `l` approaches 0 of the amount of line segments that contain at least one element of the set when you partition the number line in segments of length `l`
I think he is using cardinality. You said both their cardinalities were R. Can you bijectively map the elements of the L Cantor set and the 3L Cantor set?
@@pbj4184 I think you can. Just like how you can map the reals between 0 and 1 to the reals between 0 and 2 (by multiplying by 2), you can do the same with the cantor set. As a concrete mapping, take the number written in base 3, and multiply it by 3 (i.e. shift the decimal place). Now it covers numbers from 0 to 3 instead of 0 to 1, and all numbers are still made of 0s and 2s.
If you extend the original interval from 0..1 to 0.. 3 then you basically tripled side length. But you'd only take the section 0.. 1 and 2.. 3 so you just doubled the cantor set
This is great stuff. Hope you get much bigger in the Mathematics UA-cam sphere. This is good for showing one of the absurdities of infinite sets and real numbers. Personally, I like to refer to things like this to complain about the naming convention of "real numbers" and "imaginary numbers". The reals are completely nuts!
The length being zero isn't surprising at all if you realise how length is defined. It doesn't have much to do with the amount of points in the set. Those remain infinite throughout. It's just the sum of all the max - min of largest subsets that are continuous in nature, at each step. At first step, all points in the range [0,1] are included. So the largest subset that is continuous is [0,1] itself. 1-0 = 1. At second step, the largest subsets that are continuous are [0,1/3], [2/3,1]. 1/3 - 0 + 1 - 2/3 == 2/3. So it's no surprise that in the limiting case, when all the points become discrete and are no longer continuous, the largest length is the length of a single point itself. Which is x - x == 0.
I think your argument for uncountability is missing something. If we assumed that only the endpoints are part of the set then it would definitely be possible to enumerate them: - On the i-th iteration (with i > 0) you are adding 2^i new endpoints, which you can easily enumerate from smallest to largest. - This means that before the i-th iteration you would have 2^i points already in the set. Then you could assign labels [2^i + 0, 2^i + 1, ..., 2^i + 2^i - 1 = 2^(i+1) - 1] to the new points. It is easy to show that this would be a bijection between the endpoints and the naturals, which would mean that the set would be countable. The reason why the cantor set is uncountable is because some points that are not endpoints are also part of the set. The actual proof uses a similar idea. First you look at the expansion of the number in base 3. Just like numbers can be written in base 10 like 0.1 or 0.9562, you can think of numbers in base 3 being written as 0.0. 0.1, 0.2, 1 and so forth. A number is only ever removed if a 1 appears in the numbers after the decimal point. 0.1 in base 3, the equivalent of 2/3 in base 10, gets removed in the first iteration and you can see that in the notation. This means that the cantor set is essentially the union of all numbers with only 0s and 2s after the "decimal point". Then you can use the same argument of having infinite lists of numbers.
@@Ennar I agree that there's a bijection between the Cantor set and binary sequences of infinite size, but that's not what he showed in the video. What he showed was an injective mapping from a subset of the Cantor set (the boundaries) into the set of binary sequences of infinite size. That particular subset of the Cantor set is actually countable (see sketch of proof just above). You can't make any conclusions regarding the countability of that set from this argument alone.
@@rfMarinheiro, take a look at 7:02. Zach says: "The Cantor set is every single combination of infinite L's and R's you could possibly have," which clearly includes points that are not endpoints. He just explained what a sequence of L's and R's represents on the example of endpoints, since they are easy to visualize and never claimed that endpoints are uncountable.
@@rfMarinheiro Any point in the Cantor set can be identified with these {L,R} sequences, not just the endpoints. The endpoints in particular are identified with {L,R} sequences that are eventually constant. So, the described {L,R} sequences puts the Cantor set in bijection with binary sequences. It puts the endpoints in particular in bijection with finite binary sequences.
I came looking for this. Thanks for the explanation, I was confused because I understood that only the endpoints make it to the Cantor set, which would make it countably infinite.
But this set has to be countable if we agree on the structure of the numbers, and we can easily show that every number in this set has to, at some point in the LR representation, become stable and exclusively have L's, or exclusively R's from that point on, that position would also denote the C_n on which it was added to the set. This also means that if you attempt to create a representation as shown in the counterexample of flipping the L/R at the specific position, you would never reach a point in which the tail stabilises at exclusively L's or exclusively R's, thus the created counterexample is not in the set (by definition we would never reach the C_n in which it would be created, because at some point it would flip positions from L to R or back.
this and your last video were awesome dude -- really accessible introduction to Hausdorff's definition of dimensionality, and answered some of my musings about the Cantor set that I haven't had time to investigate. thanks!
Is this true that the Cantor set includes only multiples of powers one-third? I my understanding is that the Cantor set includes mostly irrational numbers and many rational numbers that are not multiples of integer powers of one-third. The usual example is one-fourth, which never gets removed. If it actually included only multiples of powers of one-third, there is no way it could be uncountably infinite because it would be a subset of the rationals, which are countably infinite. You'd would be able just count the rationals and skip the ones that are removed.
The Cantor set (or Cantor discontinuum) is the set of all numbers in an interval whose base-3 representation does not contain the digit 1. Then, how about the set of all numbers whose base-3 representation contains finitely many digits 1? It can be seen that this set is a union of countably many scaled down copies of the Cantor set; and so it has measure 0. (Basically: every gap in the set is filled by another copy of the Cantor set.) It's also a dense set (and, of course, uncountably infinite like Cantor set itself). I don't know if this set has a name, or what are its other topological properties.
Great video. I'd like to see another one that explores this topic even more At about 9:58 you said that the Cantor set has the same cardinality as the real numbers, but I think this statement may involve the assumption of the continuum hypothesis. You showed that the Cantor set is uncountable so it has greater cardinality than the countable rational numbers. I think the continuum hypothesis states that there are no orders of infinity between the cardinality of the integers and the real numbers, but CH has been shown to be independent of the other axioms of set theory.
The most intriguing for me is that probably infinity just exists in math but not in the real world, assuming that the space-time is discrete rather than continuous, and maybe that could be the reason of the tiny errors in many mathematical models of physical systems
Another way to break it down... If you write all of the real numbers in the interval [0,1) in base 3, they are all of the numbers of the form 0.dddddddd where each digit d is either 0 or 2. This set of numbers is isomorphic to all numbers of the form 0.dddddd where d is 0 or 1 (0 0 and 1 2 defines a unique 1-1 mapping between the sets). But the latter is the binary representation of ALL real numbers in the interval [0,1). We can toss 1 back into the sets and conclude that the Cantor set is isomorphic to the original set of real numbers on [0,1].
Something that wasn’t mentioned: if you add the cantor set C to itself, you get the interval [0,1], I.e. every number between 0 and 2 is the sum of two numbers in C!
If I understand right, the Cantor set consists of the endpoints of the line segments described in this video. They can all be expressed as fractions where the denominator is a power of 3. Therefore they are a subset of all rationals , a countably infinite set. These endpoints can be expressed as an infinite set of infinite strings of 'L"s and "R"s that contain all the possible combinations of these "L"s and "R"s. These strings can be converted to binary by converting to '1"s and '0"s giving us an infinite set of binary numbers containing all the possible combinations of binary digits. Applying Cantor's diagonal argument, we can create a new binary number that can be added to a list that already has all the possible combinations possible. But we have already shown that this set is a subset of the rationals and therefore countable. Somewhere in here there is a proof by contradiction. Am I missing something?
Yes, you are missing something. You said “the set consist of the endpoints of the line segments”, which is not true. It is true that all segment endpoints will be in the set. However, there are other numbers in it as well. Consider a number in base-3: 0.200202200220020020020… that does not terminate. It will not by an endpoint of any segment. (only the numbers that end with 000… and 222… are the endpoints)
It is almost like Infinity raised to the infinity and that pattern itself also expands infinitely! Dang. This truly gives me a whole new perspective of numbers! :)
9:10 This however works only because you have a length of the number that is equally long or longer as the entries in the list of numbers you have. So in fact, you can never find any new number that way, as you would need an infinite amount of time to create one, i.e. your attempt to create a new number would never end, and as such no new number at all would be made.
Do yourself a favor and don't try to make sense of it. The entire lecture is total nonsense, nothing can ever be proven. This guy is just goes on and on without any proof or any solid examples...pure gibberish
Here is a proof for the claim in the pinned comment. (I'm not sure how to do it by induction.) "Show that any real number in [0,2] is the sum of two elements in the Cantor set." Proof: Let p be an element of [0,2]. Take q = p/2 Let q(n) be the ternary representation of q, where n refers to decimal place (since q is less than or equal to 1, we can take the "ones" place to be 0.) Define c to be the real number represented by the same ternary sequence as q, except replace every 2 with 1. Define d to be the real number represented by the ternary sequence that has a 1 when q has a 2, and 0s everywhere else. Note that c + d = q, because when q(n) is 0 or 1, c(n) agrees and d(n) is 0. When q(n) = 2, both c(n) and d(n) are 1. Note also that since c and d are constructed so that they have only 0 and 1 in their ternary representation, 2c and 2d only have 0 and 2s in their ternary representation, hence 2c and 2d are in the Cantor set. Since c + d = q, 2c + 2d = p. Thus, the numbers a = 2c and b = 2d are in the Cantor set, and a + b = p. We win / QED
9:10 This reasoning implies that the set of all integers is also uncountably infinite, which is not true. This is because every integer can also be written as an infinite binary number. This is why I have never really accepted this reasoning. If you have an explanation to why I'm wrong, please explain it, because I would love to understand why this supposedly works.
Although you can write every integer as an infinite binary sequence (the normal binary representation with an infinite sequence of zeros in front). You can't write every infinite binary sequence as an integer. E.g. ...111 or ...101010 don't correspond to an integer
What about this, LRRLLLRRRRLLLLLRRRRRRLLLLLLLRRRRRRRRLLLLLLLLLRRRRRRRRRRLLLLLLLLLL... Alternating sequences of L's and R's and the next sequence is always one longer than the previous one.
To answer that, include M for middle as well. Then, any number in the segment [0,1] can be written as a sequence of L's, M's and R's. Of course, Cantor set is the set of those numbers in [0,1] which has no M in its representation. Now, replace L with 0, M with 1 and R with 2 as digits in ternary (base 3), and LRLRLR... becomes 0.02020202... in ternary, which can be computed as the sum of the series 0 * 1/3 + 2 * (1/3)^2 + 0 * (1/3)^3 + 2 * (1/3)^4 + ... = 1/4. See Zach's comment.
There is also a simple way to calculate it analogous to how we calculate periodic numbers in decimal. So, let me start with decimal system: 0.111111... = 1/9 = 1/(10^1-1) 0.55555... = 5/9 = 5/(10^1-1) 0.23232323... = 23/99 = 23/(10^2 - 1) 0.123123123123... = 123/999 = 123/(10^3-1) etc. Notice that you look at the length of the period and use it as exponent in the denominator. LRLRLR... is 0.020202... in ternary, so it is equal to 2/(3^2 - 1) = 2/8 = 1/4.
It's worth noting that the LRLRL... example you gave corresponds to 1/4, which is in the Cantor set. So, adding to the weirdness, although all the endpoints of the intervals have denominators that are powers of 3, in the limit, the Cantor set contains rationals whose denominators are not power of 3.
For those of you asking about the presence of irrational numbers in C, or for numbers in C that aren't endpoints, you can answer these questions using the LR sequence representation. It is in stronger analogy with decimal numbers than I think people are realizing. If we represent everything in ternary, then as done in the video, we can represent each element of the Cantor set as a sequence of {LR}, where L corresponds to 0 and R corresponds to 2. A letter for 1 isn't needed, since it never appears. Regardless of what integer base we are in, the connection between rationality and decimal representation is preserved. So: The LR sequences that are _eventually constant_ are the endpoints. If it is eventually constant with "L", we can identify those as the "terminating" decimals. If it is eventually constant with "R", it is a repeating decimal with period 1. Since the "decimal representation" is terminating/repeating, all end points are rational numbers. Furthermore, all terminating LR sequences are end points. The LR sequences that eventually repeat, but not with period 1 are rational numbers in the Cantor set that are not endpoints. Consider LRLRLR... it is (i) repeating (ii) is not eventually constant. That tells me that it is (i) a rational number (ii) not an endpoint. This sequence is identified as 1/4 in the pinned comment, which corroborates my story. And finally, any LR sequence that does not repeat nor terminate corresponds to an irrational number in the Cantor set. So, the number corresponding to LRLLRLLLRLLLLR... represents an irrational number in the Cantor Set, as does any other non-repeating/non-terminating LR sequence. Also Epstein didn't kill himself
I don't know who this Epstein is and I'm sure it's quite important to some people, but I feel disgusted by the introduction of political messages into mathematical discourse.
@@Ennar I actually thought it was really funny how he inserted a random political topic into a lengthy mathematical explanation. On that note, if you write a lengthy and helpful mathematical explanation, you definitely deserve to shamelessly plug your political opinion in there.
There is an error in the video about how to conclude the infinite size of the cantor set. The mapping of the endpoints to strings of ones and zeroes makes it a countable set which is much smaller than the set between 0 and 1. However the numbers between the endpoints remain uncountable and is therefore much larger.
it's bothering me that you can count the numbers that get into the cantor set in each step for any finite number of steps, just going through the end points, like 0; 1; 1/3; 2/3; 1/9; 2/9; 7/9; 8/9 and so on. but somehow when you're repeating that an infinite number of times, you get an uncountable amount. crazy
it's all the numbers inbetween 0 and 1, including 0 and 1, that, in ternary, don't have a 1 in them, or end with 1. ex: 1/3 is 0.1, 2/9 is 0.02, 0 is 0, and 1 is 1.
Everything in the cantor set being mappable to R is actually not that surprising: R can also be mapped to (an uncountably infinite subset of) the infinitely long binary numbers in the same way - by continually splitting a segment in half, so if you just take the cantor set instruction and reinterpret them as R-instructions, you get every real number (or, two copies of every real number)
When you showed the mapping from [0, 1] to the real numbers at 10:37 I expected that you would show the mapping from the Cantor set to [0, 1], called the Cantor function (AKA Devil's staircase). It has it's own host of interesting properties. en.wikipedia.org/wiki/Cantor_function
The mapping of the cantor set to the interval [0,1] was so counterintuitive, that I'm not convinced by it. I mean, all numbers from the cantor set are already present in the3 interval 0-1, so they can be mapped to themselves. However in the interval 0-1, apart of all numbers from the cantor set, there are also numbers, which doesn't belong to the cantor set, because they've been excluded from it. So every number on the cantor set can be mapped to itself in the interval 0-1, yet they'll remain numbers in 0-1, that would not be mapped to any number from the cantor set. What am I missing? Great video, by the way!
Yeah, you could map the elements of Cantor set to themselves in the set of [0,1], but that's not what we do in the mapping between the cantor set and [0,1]. In base 3 each number in the Cantor set is written with only 0s and 2s. Then, a mapping could be that we map any number in the cantor set to the number in [0,1] that we would get if replace all 2s with 1s and interpret it as a binary number. For example, 2/3 = 0.2000... (base 3) would be mapped to 1/2 = 0.1000... (base 2)
I'm currently at 8:05 of the video and I can't help but notice something. It's counter intuitive to say that the cantor set is a big infinity, because as far as I know, to compare infinities, you need to be able to "map" every number from one infinity to the other. You can do that for all positive integers and even positive integers, mapping goes : 0 -> 0 1 -> 2 2 -> 4 3 -> 6 4 -> 8 ... and so on. The fact that you can map those two infinities together makes them the same "size". Other example, you can map positive integers to integers, mapping goes : 0 -> 0 1 -> 1 2 -> -1 3 -> 2 4 -> -2 5 -> 3 6 -> -3 ... and so on. You couldn't, for example, map the integers to real numbers, because by the time it takes you to map reals from 0 to 1 in integers, you've ran out of integers (or at least, that's how I feel about it, there's probably a more rigorous proof out there). So what seems weird to me is that the cantor set's numbers can be mapped to integers (like binary numbers) (0...01 -> 1; 0...010 -> 2;0...011 -> 3). And the integer infinity is known to be an infinity smaller than real infinity (I think ?). So it seems weird for me that the cantor set infinity is a big one. Gotta watch the rest of the video, maybe he explains that.
If I remember something learned 20 years ago correctly, I think the Cantor set should be more "dense" than the set of interger. In your explanation of mapping infinity, the mapping use one integer to map another one. To map just one real number represented as an infinite sequence in the Cantor set, you need to use the same amount of number of integer ( which is infinity number of integer). Why the density of real number between [0, 1] is the same as the Cantor set? Because we can use different combinations of positive integers to map real, and to map Cantor set.
The truth is that Cantor set also contains infinite nonperiodic ternary numbers. These would have to be mapped to integers with actually infinite number of digits, and there are no such integers.
What’s crazy is that it is impossible to touch something knowing that to touch an object you have to move half the distance to get there than half that distance than half that distance….
Actually you can do the binary flipping thing with Integers as well. You could just add one to every digit and if you have a 9 you wrap it around to 0.
Yeah, that's usually how that's used. It's called "Cantor Diagonalization", and from the name, I would guess that Cantor used it to prove that the set of all real numbers is larger than the set of all integers.
I am absolutely scared shitless when it comes to arithmetic. I've never been good at it so never tried. As a college student with degrees I can honestly say I have calculators going back some years. Being one of the 1st in my area to carry a mobile, I did so because they had a built in calculator. So why do I love these types of video?? They more convoluted the better. I'm just starting to get it.
The video says that the Cantor set is an uncountable infinity. Why can't you simply count the entries in the Cantor set in the order they are originally created, such as 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, and so on? Doesn't this make it a countable set? If the positive integers are countable, isn't this countable in the same way? Just like you would never miss a positive integer if you count far enough, you never miss a member of the Cantor set if you count far enough.
Sometimes I like to add a dimension to sets. To see in what way they might be bigger or smaller than other sets. So, you would get the illusion of "bigger and smaller" infinities. What I like to do is to have a distance of 1 between every new line. And calculate the surface area between the new line and the previous line. What I mean is that the real numbers between 0 and 1. Will add a surface area of 1 with each new line. The Cantor set would be having a surface area of 2/3th (2 x 1/3th), then 4/9th added to it. Then 8/27th added to it. Etc. The result is that the real numbers between 0 and 1 will give a result of an infinite area. The result of the Cantor set between 0 and 1 will give a result of 3. Now, multiply the length and width by 3. The real numbers will between 0 and 3, but between the lines there will be a distance of 3. Let's say the result is now 9 times infinite (you build up to infinite, 9 times as fast). The Cantor set will... well, before we use the distance of 3 between each line. We double the cantor set. And we are adding a new surface area with a length of 2x1 (where the 1 is 3 x the 1/3th of the smaller cantor set) The previous area was 3, so we can double this and multiply it then by the distance of 3 between the lines. This gives us 18. We add that length of 2x1, multiplied by 3. Which is 6. So the total is now 24. This is 8 times bigger than the smaller version. The dimensions for the real numbers between 0 and 1 remains 2. But I find a number of 1.89279 as dimension for the cantor set. I don't know if it is a fair approach. But I like these kind of calculations any wyay. :)
ok before watching the video heres my initial thought: any number between two rational numbers gets removed eventually... given any number we should be able to come up with two rational numbers whos denominator is a power of three and gets removed eventually. so all numbers get removed? (besides endpoints)
A list of infinite numbers cannot be created because you would run out of time / memory before it can be computed upon by flipping the bit to make a new infinite number not on the set. Therefore the postulation is false.
One question I have about infinity is, if you have infinite integers (or say infinite dimensions in a vector space) can you not start to approximate real functions, in an infinite dimensional vector space each column vector could represent a close to real function if you specify how many decimal places you are using to represent say 0.00000000000001 = 1 0.0000000000002 = 2
Ok but the set is also a collection of rational numbers...how/when do they go from countable to uncountable? Later that afternoon: Correction the set of all end points are just a bunch of rationals but then you keep pushing forward (during the construction) little open intervals with points that might be in the set at the next iteration. So there are more points in those open intervals whose membership is still not known...that is the weird bit to me.
The more I think about it the less weird the cardinality seems. At each stage of the construction you have the new set of end points plus a bunch of little open intervals...there is lots of stuff in those open intervals.
Would a non-repeating infinite binary number (like e.g. 10110111011110...) be an irrational number that is part of the cantorset? It definitely would be a non-endpoint number as those would, at some point, only have 1s or 0s. Feel free to correct either statement if I'm wrong, this thought just crossed my mind.
I think what you're going for is: A non-repeating, non-terminating _ternary_ number in [0,1] without a 1 in its expansion is in the Cantor set, and it is irrational. Yes, that is true.
This one came out much longer than expected as it was a more technical video, because of that I had to omit some things I'm going to include in this comment.
1) Since the cantor set is uncountable, that means there are points in it that are NOT endpoints in any of the C_n sets (I brought this up but never acknowledged the answer). In fact, the set being uncountable means there must be irrational numbers in the set since rational numbers are countable.
2) One example of a point in the cantor set which is not an endpoint is 1/4, if you put a dot at 1/4 and move it down, it'll always be in the next C_n set but it will never be an endpoint.
3) 1/4 is known to be in the set because it has a ternary (base 3) form that does not include the number 1 (1/4 = .020202020.... in base 3). I never discussed this in the video but the cantor set consists ONLY of numbers in [0,1] that can be written in base 3 form, without the number 1. (Note: Base 2 = binary, base 3 = ternary).
4) There's a cool property about the cantor set that can be proved graphically, and if you want a challenge try to prove it. Property: For ANY number between 0 and 2 (call it p), there exists two numbers in the cantor set (call them a and b), such that a+b=p.
5) Sorry for anything that wasn't to scale, told the animator to make everything as proportional as possible but he needed room to write everything
Video published 9 minutes ago, comment a week ago :V
Facinating video, I feel like you're my math teacher! Though perhaps more interesting
@@TheSummerLab1 yeah, UA-camrs can publish and then private a video but they can comment on it
Huffman compression!
what is the value of that LRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLRLR... number?
This is what makes science and math simply unforgettable for me. To learn about things like this is magnificent and surreal. Amazing!
Just major in Math
There are aleph null numbers between [0,1]
@@mattmahoney8659if I was smarter/ had done more math when I was younger I would
@@l1mbo69 No time like the present.
One of the neat and unique things about Math is that it ages well, no single idea ever becomes obsolete as new ideas emerge unlike with Technology, medicine, and to some extent Physics and Chemistry. That's because Math eventually became based on proof and statements of truth which remain true forever. ( if proven correctly )
That's a long way of saying you can always lean it. There is nothing uniquely special about majoring in it in college anymore. To learn what Math majors know you need 2.5 things. (1) Source material like this video, (2) Practice ( old textbooks, (maybe) brilliant.org, (maybe) myopenmath courses (if that's still around) and (2.5) patience because learning math can be hard at times in a way like a foreign language or a sport can be and requires one to push through its hard parts.
In all likelihood, if your like me, you probably dont have the time/energy/resources to teach yourself all the math you wish you knew but never got to learn. But I wanted to put out there that it's not an impossible thing just in case anyone wanted to try. It's a lot easier to try something when you know its possible.
@@mattmahoney8659 thanks for the to the point structure. But I have tried, and it's not that I am terrible but I am far from good enough to justify majoring in it. I'm not going to be completely leaving it behind, though. I have decided to get a degree in Physics (and I'll try to minor in mathematics). I seem to be much better at it (judging from how I fare in international olympiads) and like it almost as much as maths.
I think the issue is that Mathematics, other than logic, also involved a lot of creativity. What is it that clicks in your brain, exactly? To me it seems to be really mysterious. All the progress I have made in maths has only been by practicing similar problems till I know the tricks used. Basically I just become familiar with that particular topic and I don't feel like I am making any progress with my core mathematical intuition. Does that make sense?
God: would you like to have length 0 or be uncountably infinite
Cantor Set: *yes*
Teacher: this exam will be straight forward.
The exam:
... is straight forward?
... but with the third and every third of every subset removed.
@@sebastianp4023 you mean every third of the exam is removed. If every third question was removed and there were n questions, what would the last question left be?
It's "straight forward" upto infinity, so the exam never ends...
Why is this lame joke on every popular math video. It's especially cringe when the math presented is not even about a problem but more of a lecture.
Fun fact: LRLRLRLR... = 1/4 so it's not just numbers that have a denominator that is a power of 3
Also, 4/9 is n the first middle third, so it’s not in their either. For people looking for a formal explanation, the set contains all number from [0,1] where the ternary (base 3) expansion has no ternary digit 1, except at the last decimal place, if it terminates.
PS, sorry if you didn’t get what I said, look it up if you really wanna know.
L, after the first iteration is the segment between 0 and 1/3, inclusive.
In the second iteration, LL would be the segment between 0 and 1/9, inclusive, and LR would be 2/9 to 1/3, inclusive.
1/4 falls on the segment between LL and LR which would be between 1/9 and 2/9, non-inclusive, and that section is removed in just the second iteration.
1/4 is NOT part of the Cantor set.
This video has a couple mistakes, but this is not one of them. All numbers in the cantor set have a denominator that is divisible by 3, with the only two exceptions being 0 and 1.
I'm open to hearing an argument that says I'm wrong and the op is right, but I'd need to see a proof.
Edit: I think the fault in your logic relies on L and R being equal to 1/2 each. Both L and R are in fact representative of 1/3 each, and a silent C is the missing variable that has been removed from the set in the middle of each L and R.
Edit #2: I was wrong. Read below if you want to know why. I'm editing my comment to note that I was wrong, but leaving it up as-is besides this note, in case others may have been mistaken, like me.
@@piezomofo I'm sorry to call you out, but when you said that 1/4 falls in between 1/9 and 2/9 it actually falls in between 2/9 and 3/9.
1/9 = 0.11111...
2/9 = 0.22222...
1/4 = 0.25
3/9 = 0.33333...
Also if you are curious about how I found that 1/4 is in the set, it has to do with this infinite sum: 1/3 - 1/9 + 1/27 - 1/81 ... = 1/4
@@samuelthecamel no, by all means, you're right to call me out. I wasn't doing the math and was quick to throw out a reply. You're right about 1/4 being between 2/9 and 3/9, and I was wrong saying it was removed at that point.
Also, I understand what you're saying about the infinite sum, but I can't wrap my head around it. I'm not saying you're wrong, in fact, it's highly likely that you're much smarter than me. I just have to imagine, though, that at some point 1/4 must be excluded because it's denominator isn't divisible by 3.
I'm probably going to drive myself nuts trying to figure this out now. What you say makes sense, but it's also contrary to this idea that the only way to end up a number in the Cantor set is to be between 0 and 1 with a denominator that's divisible by 3.
Either way, my mind has been sufficiently blown. I'll let you know if I make any progress, and thank you for correcting me. Now I've got a challenge.
@@samuelthecamel wow, I just read a paper from faculty.math.illinois.edu/~reznick/496-1-16-19.pdf It explained the same thing you said and went a little further, obviously easier to go further in a paper than in a UA-cam comment.
The paper also says that in an informal survey (sample size 12) 9 out of 12 did not know or did not remember the fact that 1/4 is in fact in the Cantor set. That made me feel a little better about being wrong, myself.
I would imagine that it stands to reason that 3/4 is also part of the set, since the set is symmetrical.
I wonder what other numbers might fall within the set that are not so obvious.
Anyway, thanks again for your original fun fact, and for setting me straight. I wouldn't have otherwise done the further research and may not have learned about this interesting non-intuitive point.
I like that this shows that a set can be uncountable and still have length 0, cause I've always wondered whether that was possible
Yep, it is the most prominent example of such a creature. In fact, coming up with such a creature without building up from the Cantor set, or its idea, is rather difficult..
@@EpicMathTime have you made any vids on this topic?
I prefer the inverse Cantor set. It has length 1, is still uncountable, and as it's length is 0+1/3+2/9+4/27... and as (1/3+2/9+4/27+..) = 1
Despite missing every number in the Cantor set, a few numbers that aren't in the inverse cantor set include 1/4 and 3/4.
@@livedandletdie i know im 10 months late, but as another comment in this comment section said, 1/4 and 3/4 are actually in the cantor set, so that is probably why the inverse cantor set does not have them
@@livedandletdie 1/4 = LRLRLRLRLRLRLRLRLRLR... and 3/4 = RLRLRLRLRLRLRLRLRLRLRLRLRLRL... in the cantor set.
Wow the intermission part really worked well
I can't believe binge watching 3Blue1Brown got me so ahead that I was completely ok with everything in this video
Same here. Im in high school watching math and physics videos on yt (3b1B, Mathloger, Numberphile, Veritasium...) just for fun and I understood everything in this video. So maybe I actually learned some math or it wasnt so complicated, but I cant tell :D
@@juliekrizkova546 same bro
Fun fact: at 7:13 you mention that you could think of the elements of the Cantor set as binary numbers, with 0's in place of the L's and 1's in place of the R's. However, if you instead replace the R's with 2's, you actually get the element of the Cantor set written in base 3. That's another way to think of how to construct the Cantor set: you take all the numbers from 0-1 written in base 3, and at each step you remove all numbers with a 1 in that digit after the decimal point.
"No one shall expel us from the Paradise that Cantor has created" - David Hilbert 1925
Cantor is really a game changer.
The best way to think about the length is that every C_n has 2/3 the length of C_n-1. If you start with 1 and multiply by 2/3 on every step, C_n will have length (2/3)^n. Since the cantor set is the limit of C_n as n approaches infinity, the length is the limit of (2/3)^n as n approaches infinity, wich is also know as zero.
BUT you are NOT removing 1/3 every time because you leave the end points. The "1/3" that you remove is smaller than each of the "1/3" segments that you remove.
@@lynnrathbun those end points have a Lebesgue measure of zero, they make no difference to the length. Just like the intervals (0, 1), (0, 1], [0, 1) and [0, 1] all have the same length of 1 unit. It's not 1- or 1+ or anything infinitesimaly close to 1, it's exactly 1. The length of an individual point is zero.
@@victorscarpes thats why a point is called dimensionless
@@stdsfromspace4560 Yes, exactly!
Or that's the cantor set only can contain points and a point is 1 dimensional essentially, thus having no length. So zero obviously comes to mind.
12:03 I feel successfully intermissed.
im jens herro
Hi Jens herro
Oh yes, time to get confused
3:17
Fun fact: The set of all fractions with denominators that are whole powers of three (i.e. the endpoints of the Cantor set) is a countable set. This means that *almost all* the points left behind are not endpoints.
Would the other numbers be non-computable numbers?
@Jorge Hernandez, all non-computable numbers are not endpoints, but there are computable non-endpoints as well.
As explained in the video, every infinite sequence of L and R describes a point in the set. All endpoints will have some initial sequence and then either infinite L's or infinite R's. Any sequence with a different end behavior, such as LRLRLR... will be a non-endpoint, and plenty of those are constructable.
This is the weird part. The set seems to be just a collection of rational numbers which is countable regardless of any other caveats. So the set is countable as you are constructing it until it hits the point at infinity then it goes uncountable?
If you're constructing it by the remove-the-middle method, all the intermediate sets are uncountable. (Any segment of non-vanishing length contains an uncountable number of points.)
@@benweieneth1103 Yea, a few minutes ago I made a comment above to that effect. At each step you get a list of end points, and some open intervals.
This testing if a number is in the cantor set reminds me of the "The terrible sound you never want to hear when working on turbine engines" meme.
I don't get why.
One takeaway from this is that "infinity" doesn't necessarily mean "all". Merely, infinity is an indicator that there is some sort of unending procedure that can be used to generate numbers, and this procedure can be used unendingly.
If you remove an infinite number of numbers between 0 and 1, there will still be infinitely many numbers left over.
And correct me if I'm wrong here, but the distance between negative infinity and positive infinity is the same as the distance from zero to infinity.
i dont think you're wrong depending on what you consider a distance, but honestly it's hard to tell because infinity is such a big number that it's hardly even a number at all. the distance between 0 and 1 is 1, and the distance between negative infinity and infinity is 2*infinity, and the distance between 0 and infinity is infinity/2. the problem of course is identifying whether any of these numbers are different at all after some point. of course, it would be incorrect to state that 1=infinity or infinity/2=infinity. it's like saying 0=1, or 1/2=1, even though 0/2=0, and 1=1. fractals are amazing
Mate, the end part on the fractional dimension was absolutely gold. I learnt the "box counting dimension" at university but it really glossed over my head when I was an undergrad. Your analogy is so simple to understand, that I'll probably use it to explain it to my students. Thank you :)
9:00
What baffles me is you could do the same thing with infinitely long decimal integers and discover that there are uncountably many of them.
Despite the fact that there are countably many decimal integers total.
The problem then is that you'd get a real number which actually isn't an integer (or fraction if you will) thus not in that set, so you haven't reached a contradiction with this new number you made
The reason this doesn't work with the integers is that every integer has a finite length. The set of integers contains elements that are arbitrarily large, but every individual element is finite in length.
If you try to apply the diagonalization method with the integers, you will end up with a number of infinite length, so it won't be a valid integer.
@@kaiblack4489
Yeah I don't know what I was thinking to be honest. Me 3 years ago not as smart as me now I guess (hopefully at least).
What I described appears to be akin to 10-adic numbers, which is a representation of reals, not just integers.
you are only the second one that explained the "I can always find a number thats not on your list" in an understandable way, thanks!
It is easy to find a number written in the LRLR form. Just replace all the L's with 0's and all the R's with 2's, and you will get the number in base-3.
For example, 1/3 = LRRRRRRRRRRRRRRR = 0.0222222222222222 (base-3) = 0.33333333333 (base-10)
You could also go the other way around to find the LRLR form from the number.
For example, 1/4 = 0.25 (base-10) = 0.02020202020202020202... (base-3) = LRLRLRLRLRLRLRLRLRLR... in the cantor set.
Why use base 3 instead of base 2? Actually you have to use base 2 instead of 3 otherwise you can't represent numbers like 4/9 which in base 3 is 0.11 or 0.102222222 both of which contains '1', but since you can't represent 1 in your method you can't represent 4/9. If you changed to base 2 it would solve that problem (i.e. L=0,R=1 like he did in the video).
@@tomerwolberg37 But that's exactly what makes it convenient. Any number which has a '1' in its base-3 representation will never be in the cantor set (4/9 for example is not in the set).
It is also possible that it is an endpoint in the set (e.g. 1/3 is 0.1 in base-3, but the convenient part is that it can be written as 0.02222... which makes it a part of the set as an endpoint).
@@karan_jain oh, I thought you were trying to show it's the same size as the real numbers. Now I see what you mean, clever.
karan jan thanks for your example, yet i am a bit lost and because you clearly have a grasp of deep mathematics maybe you could help. what is this all about? Why was Cantor showing this "set"? I ask because i can somewhat follow along witht this video, but I dont know what its all about. Why are we talking about this set? What is its significance? I know its EXTREMELY significant, yet as I am ignorant i cant figure out why. Maybe you would have interest answering?
One curious thing about the Cantor set, is that, yes, it contains all the endpoints of the C0, C1, C2, ... sequence, but it does not ONLY contain those endpoints. This is clear, because there are countably many such endpoints---they can be enumerated. The Cantor set contains uncountably many points which are not endpoints of any such interval.
I read about this in Chaos: a new science. Nice!
Bro, youre a hell of a teacher. I listen to your videos while driving (being a father and husband really limits the time frame i have to focus on myself, so i do what i can) and despite that, im able to grasp everything youre talking about clearly. I wish i had a teacher that was as good as you are, back when i was in school, maybe i would have pursued higher education.
I've just invented the non-inclusive cantor set, where instead of including the endpoints you exclude them.
The full list of non-inclusive cantor numbers is as follows:
One year late: 1/4
I just was researching fractional dimensions this morning, and this video comes out no more than 6 hours later. What a coincidence
not likely. abc corp is just really good at tracking your behavior.
The algorithm is doing a good job
"The size of the Cantor set gets doubled when its side length is tripled." Could you please elaborate on how do you define double here? If you use cardinality, it is always that of R; if you use Lebesgue measure, it is always 0.
Having seen the video by 3blue1brown he references, I think that the size he is talking about is defined as the limit as `l` approaches 0 of the amount of line segments that contain at least one element of the set when you partition the number line in segments of length `l`
I think he is using cardinality. You said both their cardinalities were R. Can you bijectively map the elements of the L Cantor set and the 3L Cantor set?
@@pbj4184 I think you can. Just like how you can map the reals between 0 and 1 to the reals between 0 and 2 (by multiplying by 2), you can do the same with the cantor set.
As a concrete mapping, take the number written in base 3, and multiply it by 3 (i.e. shift the decimal place). Now it covers numbers from 0 to 3 instead of 0 to 1, and all numbers are still made of 0s and 2s.
@@amaarquadri You are correct. I wonder what he meant then 🤔
If you extend the original interval from 0..1 to 0.. 3 then you basically tripled side length. But you'd only take the section 0.. 1 and 2.. 3 so you just doubled the cantor set
This is great stuff. Hope you get much bigger in the Mathematics UA-cam sphere. This is good for showing one of the absurdities of infinite sets and real numbers. Personally, I like to refer to things like this to complain about the naming convention of "real numbers" and "imaginary numbers". The reals are completely nuts!
I doubt 0 has more L’s than my life
This video should get love as much as the size of cantor set...❤
The length being zero isn't surprising at all if you realise how length is defined. It doesn't have much to do with the amount of points in the set. Those remain infinite throughout. It's just the sum of all the max - min of largest subsets that are continuous in nature, at each step. At first step, all points in the range [0,1] are included. So the largest subset that is continuous is [0,1] itself. 1-0 = 1.
At second step, the largest subsets that are continuous are [0,1/3], [2/3,1]. 1/3 - 0 + 1 - 2/3 == 2/3.
So it's no surprise that in the limiting case, when all the points become discrete and are no longer continuous, the largest length is the length of a single point itself. Which is x - x == 0.
Thanks for being an awesome channel
I think your argument for uncountability is missing something. If we assumed that only the endpoints are part of the set then it would definitely be possible to enumerate them:
- On the i-th iteration (with i > 0) you are adding 2^i new endpoints, which you can easily enumerate from smallest to largest.
- This means that before the i-th iteration you would have 2^i points already in the set. Then you could assign labels [2^i + 0, 2^i + 1, ..., 2^i + 2^i - 1 = 2^(i+1) - 1] to the new points.
It is easy to show that this would be a bijection between the endpoints and the naturals, which would mean that the set would be countable. The reason why the cantor set is uncountable is because some points that are not endpoints are also part of the set.
The actual proof uses a similar idea. First you look at the expansion of the number in base 3. Just like numbers can be written in base 10 like 0.1 or 0.9562, you can think of numbers in base 3 being written as 0.0. 0.1, 0.2, 1 and so forth. A number is only ever removed if a 1 appears in the numbers after the decimal point. 0.1 in base 3, the equivalent of 2/3 in base 10, gets removed in the first iteration and you can see that in the notation. This means that the cantor set is essentially the union of all numbers with only 0s and 2s after the "decimal point". Then you can use the same argument of having infinite lists of numbers.
Zach's argument is simply that there is a bijection between elements of Cantor set and binary sequences and 2^(aleph_0) is uncountable.
@@Ennar I agree that there's a bijection between the Cantor set and binary sequences of infinite size, but that's not what he showed in the video. What he showed was an injective mapping from a subset of the Cantor set (the boundaries) into the set of binary sequences of infinite size. That particular subset of the Cantor set is actually countable (see sketch of proof just above). You can't make any conclusions regarding the countability of that set from this argument alone.
@@rfMarinheiro, take a look at 7:02.
Zach says: "The Cantor set is every single combination of infinite L's and R's you could possibly have," which clearly includes points that are not endpoints. He just explained what a sequence of L's and R's represents on the example of endpoints, since they are easy to visualize and never claimed that endpoints are uncountable.
@@rfMarinheiro Any point in the Cantor set can be identified with these {L,R} sequences, not just the endpoints. The endpoints in particular are identified with {L,R} sequences that are eventually constant.
So, the described {L,R} sequences puts the Cantor set in bijection with binary sequences. It puts the endpoints in particular in bijection with finite binary sequences.
I came looking for this. Thanks for the explanation, I was confused because I understood that only the endpoints make it to the Cantor set, which would make it countably infinite.
11:50, never been mad before for having UA-cam premium
I didn't get an ad either, no premium. Hope that makes you feel better :)
But this set has to be countable if we agree on the structure of the numbers, and we can easily show that every number in this set has to, at some point in the LR representation, become stable and exclusively have L's, or exclusively R's from that point on, that position would also denote the C_n on which it was added to the set. This also means that if you attempt to create a representation as shown in the counterexample of flipping the L/R at the specific position, you would never reach a point in which the tail stabilises at exclusively L's or exclusively R's, thus the created counterexample is not in the set (by definition we would never reach the C_n in which it would be created, because at some point it would flip positions from L to R or back.
this and your last video were awesome dude -- really accessible introduction to Hausdorff's definition of dimensionality, and answered some of my musings about the Cantor set that I haven't had time to investigate. thanks!
I’m a math teacher who spent way too long in college and graduate school. Congratulations, you finally got me to care about the Cantor set!! 🎉
Is this true that the Cantor set includes only multiples of powers one-third? I my understanding is that the Cantor set includes mostly irrational numbers and many rational numbers that are not multiples of integer powers of one-third. The usual example is one-fourth, which never gets removed. If it actually included only multiples of powers of one-third, there is no way it could be uncountably infinite because it would be a subset of the rationals, which are countably infinite. You'd would be able just count the rationals and skip the ones that are removed.
The Cantor set (or Cantor discontinuum) is the set of all numbers in an interval whose base-3 representation does not contain the digit 1. Then, how about the set of all numbers whose base-3 representation contains finitely many digits 1? It can be seen that this set is a union of countably many scaled down copies of the Cantor set; and so it has measure 0. (Basically: every gap in the set is filled by another copy of the Cantor set.) It's also a dense set (and, of course, uncountably infinite like Cantor set itself).
I don't know if this set has a name, or what are its other topological properties.
I remember seeing this is college. Fun times!
Great video. I'd like to see another one that explores this topic even more At about 9:58 you said that the Cantor set has the same cardinality as the real numbers, but I think this statement may involve the assumption of the continuum hypothesis. You showed that the Cantor set is uncountable so it has greater cardinality than the countable rational numbers. I think the continuum hypothesis states that there are no orders of infinity between the cardinality of the integers and the real numbers, but CH has been shown to be independent of the other axioms of set theory.
We spend 1/3 of our lives sleeping, and the other 2/3 in the dark.
The most intriguing for me is that probably infinity just exists in math but not in the real world, assuming that the space-time is discrete rather than continuous, and maybe that could be the reason of the tiny errors in many mathematical models of physical systems
Another way to break it down... If you write all of the real numbers in the interval [0,1) in base 3, they are all of the numbers of the form 0.dddddddd where each digit d is either 0 or 2. This set of numbers is isomorphic to all numbers of the form 0.dddddd where d is 0 or 1 (0 0 and 1 2 defines a unique 1-1 mapping between the sets). But the latter is the binary representation of ALL real numbers in the interval [0,1). We can toss 1 back into the sets and conclude that the Cantor set is isomorphic to the original set of real numbers on [0,1].
Something that wasn’t mentioned: if you add the cantor set C to itself, you get the interval [0,1], I.e. every number between 0 and 2 is the sum of two numbers in C!
Me at 2 Am: Casually studying about how Cantor Set has log_3(2) dimension
All infinites are equal, but some infinites are more equal than others.
If I understand right, the Cantor set consists of the endpoints of the line segments described in this video. They can all be expressed as fractions where the denominator is a power of 3. Therefore they are a subset of all rationals , a countably infinite set. These endpoints can be expressed as an infinite set of infinite strings of 'L"s and "R"s that contain all the possible combinations of these "L"s and "R"s. These strings can be converted to binary by converting to '1"s and '0"s giving us an infinite set of binary numbers containing all the possible combinations of binary digits. Applying Cantor's diagonal argument, we can create a new binary number that can be added to a list that already has all the possible combinations possible. But we have already shown that this set is a subset of the rationals and therefore countable. Somewhere in here there is a proof by contradiction. Am I missing something?
Yes, you are missing something. You said “the set consist of the endpoints of the line segments”, which is not true.
It is true that all segment endpoints will be in the set.
However, there are other numbers in it as well. Consider a number in base-3:
0.200202200220020020020… that does not terminate. It will not by an endpoint of any segment.
(only the numbers that end with 000… and 222… are the endpoints)
I had never heard of fractional dimensions before. Mind expanded. Thanks!
when a is an integer, a=3^n, and m
It is almost like Infinity raised to the infinity and that pattern itself also expands infinitely! Dang. This truly gives me a whole new perspective of numbers! :)
This is insane. Fractional dimensions? I’ve never even heard about that before!
relevant search term: Hausdorff Dimension
two years late but oh well
@@wyboo2019 I’ll look that up. Thanks
9:10 This however works only because you have a length of the number that is equally long or longer as the entries in the list of numbers you have. So in fact, you can never find any new number that way, as you would need an infinite amount of time to create one, i.e. your attempt to create a new number would never end, and as such no new number at all would be made.
Do yourself a favor and don't try to make sense of it. The entire lecture is total nonsense, nothing can ever be proven.
This guy is just goes on and on without any proof or any solid examples...pure gibberish
Here is a proof for the claim in the pinned comment. (I'm not sure how to do it by induction.) "Show that any real number in [0,2] is the sum of two elements in the Cantor set."
Proof:
Let p be an element of [0,2]. Take q = p/2
Let q(n) be the ternary representation of q, where n refers to decimal place (since q is less than or equal to 1, we can take the "ones" place to be 0.)
Define c to be the real number represented by the same ternary sequence as q, except replace every 2 with 1.
Define d to be the real number represented by the ternary sequence that has a 1 when q has a 2, and 0s everywhere else.
Note that c + d = q, because when q(n) is 0 or 1, c(n) agrees and d(n) is 0. When q(n) = 2, both c(n) and d(n) are 1.
Note also that since c and d are constructed so that they have only 0 and 1 in their ternary representation, 2c and 2d only have 0 and 2s in their ternary representation, hence 2c and 2d are in the Cantor set.
Since c + d = q, 2c + 2d = p.
Thus, the numbers a = 2c and b = 2d are in the Cantor set, and a + b = p.
We win / QED
Ah yes, your classic "and we win". Nice way to do it, exploiting the properties of the Cantor set that arise from using ternary numbers
9:10 This reasoning implies that the set of all integers is also uncountably infinite, which is not true. This is because every integer can also be written as an infinite binary number. This is why I have never really accepted this reasoning. If you have an explanation to why I'm wrong, please explain it, because I would love to understand why this supposedly works.
Although you can write every integer as an infinite binary sequence (the normal binary representation with an infinite sequence of zeros in front). You can't write every infinite binary sequence as an integer. E.g. ...111 or ...101010 don't correspond to an integer
What would be the actual number that corresponds to LRLRLRLRLRLRLR... in the Cantor set?
2/7 I think? You can find that through a geometric series
What about this,
LRRLLLRRRRLLLLLRRRRRRLLLLLLLRRRRRRRRLLLLLLLLLRRRRRRRRRRLLLLLLLLLL...
Alternating sequences of L's and R's and the next sequence is always one longer than the previous one.
To answer that, include M for middle as well. Then, any number in the segment [0,1] can be written as a sequence of L's, M's and R's. Of course, Cantor set is the set of those numbers in [0,1] which has no M in its representation. Now, replace L with 0, M with 1 and R with 2 as digits in ternary (base 3), and LRLRLR... becomes 0.02020202... in ternary, which can be computed as the sum of the series 0 * 1/3 + 2 * (1/3)^2 + 0 * (1/3)^3 + 2 * (1/3)^4 + ... = 1/4. See Zach's comment.
There is also a simple way to calculate it analogous to how we calculate periodic numbers in decimal.
So, let me start with decimal system:
0.111111... = 1/9 = 1/(10^1-1)
0.55555... = 5/9 = 5/(10^1-1)
0.23232323... = 23/99 = 23/(10^2 - 1)
0.123123123123... = 123/999 = 123/(10^3-1)
etc.
Notice that you look at the length of the period and use it as exponent in the denominator.
LRLRLR... is 0.020202... in ternary, so it is equal to 2/(3^2 - 1) = 2/8 = 1/4.
I think I just wooshed myself
Coming to the realization that each split of the cantor set becomes its own cantor set without him having to tell me was a fun thing
It's worth noting that the LRLRL... example you gave corresponds to 1/4, which is in the Cantor set. So, adding to the weirdness, although all the endpoints of the intervals have denominators that are powers of 3, in the limit, the Cantor set contains rationals whose denominators are not power of 3.
Awesome explanation! Thanks.
I have watched both vsauce and numberphile. I personally like your channel better. Keep up the good work!
Weird things happen when you go beyond countable infinity.
Weird things happen when you get to countable infinity.
Weird things happen when you turn on your wifi.
Aleph null,omega omega+1....
Sounds like somebody's taking Intro to the Theory of Computation this semester lol (such as CS520 if you go to UW-Madison)
What happens at infinity? NOTHING, BECAUSE YOU CAN NOT GET THERE.
This is a wonderful video. Well-done.
For those of you asking about the presence of irrational numbers in C, or for numbers in C that aren't endpoints, you can answer these questions using the LR sequence representation. It is in stronger analogy with decimal numbers than I think people are realizing.
If we represent everything in ternary, then as done in the video, we can represent each element of the Cantor set as a sequence of {LR}, where L corresponds to 0 and R corresponds to 2. A letter for 1 isn't needed, since it never appears.
Regardless of what integer base we are in, the connection between rationality and decimal representation is preserved. So:
The LR sequences that are _eventually constant_ are the endpoints. If it is eventually constant with "L", we can identify those as the "terminating" decimals. If it is eventually constant with "R", it is a repeating decimal with period 1. Since the "decimal representation" is terminating/repeating, all end points are rational numbers. Furthermore, all terminating LR sequences are end points.
The LR sequences that eventually repeat, but not with period 1 are rational numbers in the Cantor set that are not endpoints. Consider LRLRLR... it is (i) repeating (ii) is not eventually constant. That tells me that it is (i) a rational number (ii) not an endpoint. This sequence is identified as 1/4 in the pinned comment, which corroborates my story.
And finally, any LR sequence that does not repeat nor terminate corresponds to an irrational number in the Cantor set.
So, the number corresponding to LRLLRLLLRLLLLR... represents an irrational number in the Cantor Set, as does any other non-repeating/non-terminating LR sequence.
Also Epstein didn't kill himself
😂
I don't know who this Epstein is and I'm sure it's quite important to some people, but I feel disgusted by the introduction of political messages into mathematical discourse.
@@Ennar I actually thought it was really funny how he inserted a random political topic into a lengthy mathematical explanation. On that note, if you write a lengthy and helpful mathematical explanation, you definitely deserve to shamelessly plug your political opinion in there.
@@Ennar That's just how I sign my mathematical discourse, dude.
@@vojtechstrnad1 you are free to feel that way, just as I am free to feel whatever I feel.
There is an error in the video about how to conclude the infinite size of the cantor set. The mapping of the endpoints to strings of ones and zeroes makes it a countable set which is much smaller than the set between 0 and 1. However the numbers between the endpoints remain uncountable and is therefore much larger.
Fun fact: the 2019 Paper 1 Mathematics Leaving Cert Exam (Higher Level) has a question all about the cantor set. It was Question 7.
it's bothering me that you can count the numbers that get into the cantor set in each step for any finite number of steps, just going through the end points, like 0; 1; 1/3; 2/3; 1/9; 2/9; 7/9; 8/9 and so on. but somehow when you're repeating that an infinite number of times, you get an uncountable amount. crazy
it's all the numbers inbetween 0 and 1, including 0 and 1, that, in ternary, don't have a 1 in them, or end with 1. ex: 1/3 is 0.1, 2/9 is 0.02, 0 is 0, and 1 is 1.
Lovely!! Thanks, Zach.
Everything in the cantor set being mappable to R is actually not that surprising:
R can also be mapped to (an uncountably infinite subset of) the infinitely long binary numbers in the same way - by continually splitting a segment in half, so if you just take the cantor set instruction and reinterpret them as R-instructions, you get every real number (or, two copies of every real number)
When you showed the mapping from [0, 1] to the real numbers at 10:37 I expected that you would show the mapping from the Cantor set to [0, 1], called the Cantor function (AKA Devil's staircase). It has it's own host of interesting properties.
en.wikipedia.org/wiki/Cantor_function
The mapping of the cantor set to the interval [0,1] was so counterintuitive, that I'm not convinced by it. I mean, all numbers from the cantor set are already present in the3 interval 0-1, so they can be mapped to themselves. However in the interval 0-1, apart of all numbers from the cantor set, there are also numbers, which doesn't belong to the cantor set, because they've been excluded from it. So every number on the cantor set can be mapped to itself in the interval 0-1, yet they'll remain numbers in 0-1, that would not be mapped to any number from the cantor set. What am I missing? Great video, by the way!
Yeah, you could map the elements of Cantor set to themselves in the set of [0,1], but that's not what we do in the mapping between the cantor set and [0,1]. In base 3 each number in the Cantor set is written with only 0s and 2s. Then, a mapping could be that we map any number in the cantor set to the number in [0,1] that we would get if replace all 2s with 1s and interpret it as a binary number. For example, 2/3 = 0.2000... (base 3) would be mapped to 1/2 = 0.1000... (base 2)
Your videos are so underrated.
I'm currently at 8:05 of the video and I can't help but notice something. It's counter intuitive to say that the cantor set is a big infinity, because as far as I know, to compare infinities, you need to be able to "map" every number from one infinity to the other. You can do that for all positive integers and even positive integers, mapping goes :
0 -> 0
1 -> 2
2 -> 4
3 -> 6
4 -> 8
... and so on.
The fact that you can map those two infinities together makes them the same "size". Other example, you can map positive integers to integers, mapping goes :
0 -> 0
1 -> 1
2 -> -1
3 -> 2
4 -> -2
5 -> 3
6 -> -3
... and so on.
You couldn't, for example, map the integers to real numbers, because by the time it takes you to map reals from 0 to 1 in integers, you've ran out of integers (or at least, that's how I feel about it, there's probably a more rigorous proof out there).
So what seems weird to me is that the cantor set's numbers can be mapped to integers (like binary numbers) (0...01 -> 1; 0...010 -> 2;0...011 -> 3). And the integer infinity is known to be an infinity smaller than real infinity (I think ?). So it seems weird for me that the cantor set infinity is a big one.
Gotta watch the rest of the video, maybe he explains that.
If I remember something learned 20 years ago correctly, I think the Cantor set should be more "dense" than the set of interger. In your explanation of mapping infinity, the mapping use one integer to map another one. To map just one real number represented as an infinite sequence in the Cantor set, you need to use the same amount of number of integer ( which is infinity number of integer). Why the density of real number between [0, 1] is the same as the Cantor set? Because we can use different combinations of positive integers to map real, and to map Cantor set.
The truth is that Cantor set also contains infinite nonperiodic ternary numbers. These would have to be mapped to integers with actually infinite number of digits, and there are no such integers.
@@u.v.s.5583 Right. That makes sense
What’s crazy is that it is impossible to touch something knowing that to touch an object you have to move half the distance to get there than half that distance than half that distance….
12:00 That's really nice man, I appreciate it
Actually you can do the binary flipping thing with Integers as well. You could just add one to every digit and if you have a 9 you wrap it around to 0.
Yeah, that's usually how that's used. It's called "Cantor Diagonalization", and from the name, I would guess that Cantor used it to prove that the set of all real numbers is larger than the set of all integers.
I am absolutely scared shitless when it comes to arithmetic. I've never been good at it so never tried. As a college student with degrees I can honestly say I have calculators going back some years. Being one of the 1st in my area to carry a mobile, I did so because they had a built in calculator. So why do I love these types of video?? They more convoluted the better. I'm just starting to get it.
The video says that the Cantor set is an uncountable infinity. Why can't you simply count the entries in the Cantor set in the order they are originally created, such as 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, and so on? Doesn't this make it a countable set? If the positive integers are countable, isn't this countable in the same way? Just like you would never miss a positive integer if you count far enough, you never miss a member of the Cantor set if you count far enough.
I got a dominos ad right after staring at the screen while zooming into the cantor set and now I really want pizza.
Sometimes I like to add a dimension to sets. To see in what way they might be bigger or smaller than other sets. So, you would get the illusion of "bigger and smaller" infinities.
What I like to do is to have a distance of 1 between every new line.
And calculate the surface area between the new line and the previous line.
What I mean is that the real numbers between 0 and 1. Will add a surface area of 1 with each new line.
The Cantor set would be having a surface area of 2/3th (2 x 1/3th), then 4/9th added to it. Then 8/27th added to it. Etc.
The result is that the real numbers between 0 and 1 will give a result of an infinite area.
The result of the Cantor set between 0 and 1 will give a result of 3.
Now, multiply the length and width by 3.
The real numbers will between 0 and 3, but between the lines there will be a distance of 3. Let's say the result is now 9 times infinite (you build up to infinite, 9 times as fast).
The Cantor set will... well, before we use the distance of 3 between each line.
We double the cantor set. And we are adding a new surface area with a length of 2x1 (where the 1 is 3 x the 1/3th of the smaller cantor set)
The previous area was 3, so we can double this and multiply it then by the distance of 3 between the lines. This gives us 18.
We add that length of 2x1, multiplied by 3. Which is 6.
So the total is now 24. This is 8 times bigger than the smaller version.
The dimensions for the real numbers between 0 and 1 remains 2.
But I find a number of 1.89279 as dimension for the cantor set.
I don't know if it is a fair approach. But I like these kind of calculations any wyay. :)
very neat introduction to the properties of the Cantor set. wish it had existed when I was learning analysis for the first time.
When we needed him most, he returned
Your header, nice bro😂
"(...) Only the avatar, master of all four elements, could stop them. But when we needed him the most, he vanished.(...)"
ok before watching the video heres my initial thought: any number between two rational numbers gets removed eventually... given any number we should be able to come up with two rational numbers whos denominator is a power of three and gets removed eventually. so all numbers get removed? (besides endpoints)
A list of infinite numbers cannot be created because you would run out of time / memory before it can be computed upon by flipping the bit to make a new infinite number not on the set. Therefore the postulation is false.
Please.. Continue your efforts to increase curiosity in every student about science and math...I thank UA-cam for recommending your channel😍
Why did the youtube algorithm gods hide this channel from me for so long?
Amazing content my dude!
One question I have about infinity is, if you have infinite integers (or say infinite dimensions in a vector space) can you not start to approximate real functions, in an infinite dimensional vector space each column vector could represent a close to real function if you specify how many decimal places you are using to represent say 0.00000000000001 = 1 0.0000000000002 = 2
Ok but the set is also a collection of rational numbers...how/when do they go from countable to uncountable?
Later that afternoon: Correction the set of all end points are just a bunch of rationals but then you keep pushing forward (during the construction) little open intervals with points that might be in the set at the next iteration. So there are more points in those open intervals whose membership is still not known...that is the weird bit to me.
The more I think about it the less weird the cardinality seems. At each stage of the construction you have the new set of end points plus a bunch of little open intervals...there is lots of stuff in those open intervals.
"it is not obvious if there are non end points at the end" - what end? if you continue indefinitely then by definition there is no end :)
Would a non-repeating infinite binary number (like e.g. 10110111011110...) be an irrational number that is part of the cantorset?
It definitely would be a non-endpoint number as those would, at some point, only have 1s or 0s.
Feel free to correct either statement if I'm wrong, this thought just crossed my mind.
I think what you're going for is: A non-repeating, non-terminating _ternary_ number in [0,1] without a 1 in its expansion is in the Cantor set, and it is irrational. Yes, that is true.
@@EpicMathTime Thank you, found a proof today as well
Something that is both 0 and uncountably infinite... seems like the kind of thing that would be in JoJo
Finally! Someone that can include their mid roll ad spam at a break point that makes some sense
What happens at infinity?
The possibilities are endless.
What would the shape of the universe be if you could map the current density function inside a single electron?
what about making an infinite zeros to the right and counting 1+1 to the right in a binary mode? wouldn't it make tha cantor set countable?
if a set has dimension n
When you showed a hyperbolic grap, it all had become clear to me. Holy shit, (0,1) is really the same size as (neg. Inf, pos. Inf)
Zach star “0 = LLLL...” me an intellectual so 0 = infinite Luigi
I went to Curiosity Stream (I'm a longtime subscriber) but I couldn't find anything titled, "The Secret Life of Chaos"