The Devil's Staircase | Infinite Series
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- Опубліковано 18 тра 2017
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Topology Riddles
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Written and Hosted by Kelsey Houston-Edwards
Produced by Rusty Ward
Graphics by Ray Lux
Made by Kornhaber Brown (www.kornhaberbrown.com)
Cantor’s Function, also known as the Devil’s Staircase, is exceptionally strange. After mapping out Cantor’s Function, named after Georg Cantor, we find that its derivative disappears almost everywhere.
Challenge Winner:
Justin Wieland
Honorable Mentions:
Felix Beichler
Andrew Weller
Comments answered by Kelsey:
Thomas Perry
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Ralph Dratman
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Math is awesome when it's not mandatory xD
Ye
This is the realest thing
The way she explained what a base is and how it works is outright brilliant.
Quickest way to prove the Cantor set’s uncountability: Every member of the set (represented in base 3), maps to a base 2 number. Just switch out the 2’s for 1’s.
Lol, I thought the proof was very complex. That's neat
Neat. A surjective map from Cantor to the unit interval.
@@SaveSoilSaveSoil (actually pretty sure it's bijective)
@@tsawy6 She did not say it's *not* injective, so she did not say anything wrong.
@@ShehabEllithy Sure, I wasn't tryna correct so much as expand
What do you play at the top of the Devil's Staircase?
Gabriel's Horn.
I actually play the horn, but my name isn't Gabriel.
For those who may not know, "Gabrieel's Horn" is a mathamatical construction with finite volume and infinite sueface area. See WIKIPEDIA
bockmaker good one
you play Rocky's theme song
Ilm Brk
If you don't get it, Gabriel's Horn is a weird object also a fractal
Hey guys when you put a red bar across the bottom of the thumbnail it looks like I have already watched the video. Maybe use a non red color for the border so people don't miss videos thinking they already watched it.
Actually there's this really cool things that all humans are equipped with called memory!
GunsExplosivesetc.
Yeah, I've memorized all my contacts' phone numbers, addresses, birthdays, favorite sushi's and colors...
In between the hundreds of clickbaity titles filling up our UA-cam feeds we frequently see titles that we think we want to watch but don't have time to watch until later. After seeing a particular title and thumbnail often enough the flaws within human memory mean that it becomes rather uncertain whether we watched a particular video or not. That's why all youtube users have this really cool feature that shows them previously watched videos when they are logged in.
It's a strategy to get more views. The red bar that covers the place where UA-cam's red bar would be so that you can't be sure if you watched the video, so you watch it again to be sure.
@@snbeast9545 sure, because it's not like they do that bar thing with every video, and this time it just so happened to be red...
Cantor's Set is my second favorite set....
...definitely top 3 :D
Answer: Cantor set in base 3 is equivalent to binary in it's size; binary is uncountably infinite
Simple proof of the uncountability of the Cantor set:
Notice that the only digits of numbers in the Cantor set are 0 and 2. The Cantor set can therefore have a one-to-one map to all the binary numbers between zero and one, simply by replacing all the 2s of the base-three-Cantor-set numbers with 1s to give base-two numbers.† As the Cantor set is all the possible permutations of 0 and 2, the corresponding binary set will perfectly span the interval from zero to one. That interval has previously been proved to be uncountable,* therefore so is the Cantor set. _Q.e.d._
* Notably by Cantor himself, in his Diagonal Argument. Thank you, past mathematicians.
† I have been corrected on this point. This is no one-to-one map, as it takes both ⅓ and ⅔ to ½. However, we can note that it is a many-to-one‡ from the Cantor set to the binary set, therefore the Cantor set is bigger,¶ and therefore the proof stands.
‡ Correction: it is a surjection.
¶ Or the same size.
Edit: spelling.
Edit2: But I have a question. Is the Cantor function a one-to-one function ? Does it have an infinite slope at points in the Cantor set ? I would be very impressed if it is and doesn't. Edit3: Query answered. It wasn't a particularly clever query.
Asthmen according to my browser anyway, yours is the first proof. Seems right to
sort by "newest first", scroll down, yes its first.
The Cantor function is surjective, but not injective so it can't be one to one. I'm not sure if there exists a one to one function, but there can be no continuous bijection because the cantor set is not connected, while [0, 1] is connected.
The map you defined is not one-to-one. It takes 1/3 and 2/3 to 1/2. Your proof still works though, you just have to fix injective to surjective.
For the Cantor set (in base 3) you will keep 0, 1 and 0, 01, ... I guess you mean all numbers whose *infinite* expansions do not contain 1, and you write 0,1 as 0,022222.... (It is also obvious that some points have to remain, as the intervals are closed. It remains amazing that the set remains uncountable, but obvious by diagonalisation). Nice video!
The Cantor Diagonalization argument would be the classic way to show uncountability. To make things more fun, I'll try to show uncountability in a different way. I will take for granted the facts that (1) the nonnegative real numbers are uncountable and (2) one can construct a convergent series to any positive real number using only the terms of the form 1/n (n a positive integer) with each term appearing only once.
Now for each number in the Cantor set, we can define an infinite series as follows: consider the base three expansion of the number. If the nth digit of the expansion is 0, then the nth digit of the series is 0. If the nth digit of the expansion is 2, then the nth digit of the series is 1/n*. For example,
0.02202... -> 0 + 1/2 + 1/3 + 0 + 1/5 ...
0.22202... -> 1 + 1/2 + 1/3 + 0 + 1/5 ...
Now for each number x in the Cantor set, we can define the following mapping:
If the corresponding series of x converges, map x to that sum. Otherwise, map x to zero.
This mapping takes numbers in the Cantor set and gives us a nonnegative real number. Not only that, but the mapping should hit *every* nonnegative real number by our assumption (2). This means that there must be at least as many numbers in the Cantor set as there are in the nonnegative real numbers. By assumption (1), there are uncountably many nonnegative reals, so there are uncountably many numbers in the Cantor set.
* For numbers in the Cantor set whose base 3 expansion terminates with a 1, for instance 0.1, we can use the alternate expansion of 0.0222... and apply the mapping to that.
Nice proof!
Cantor's set is uncountable because there is an isomorphism from the base 3 representation of its numbers to the binary representation of the real numbers of the interval [0,1] and this interval is known to be uncountable. One example of such isomorphism consists of replacing the 2s by 1s.
I can't stress how important this series is. Execution is brilliant and I wish more people can appreciate the content.
I love your series! Thank you for releasing this and all of your videos!
Assume Cantor Set is countable, which also means listable.
Write down all possible members of the set in base 3.
1: 0.20222220000202....
2: 0.22220020202020....
3: 0.20202220202022....
etc.
Now for each member evaluate the decimal place which corresponds with its position in the list (the first point of the first member, etc.).
Now create a new number by using the opposite of the point we just found (switch 2's and 0's from first).
This number then must be different from every number contained within the list.
Therefore the list is incomplete.
Therefore the Cantor Set is uncountable.
Sorry, but I don't see how this holds against my counterexample showing that the Cantor set is in fact _countable_ :
#1: 0
#2: 1
#3: 0.1
#4: 0.2
#5: 0.01
#6: 0.02
#7: 0.21
#8: 0.22
#9: 0.001
...
Continuing after this pattern, you can count every number in the Cantor set. To test that, just give me a number in the set and it will be in this list.
Did you just use the Cantor diagonal argument over the Cantor set? Cantor-ception!
Iwer Sonsch your map misses all infinitely long decimals. Well not decimals since it's base 3, I guess trimals? Regardless, your map misses all infinitely long representations. The diagonalization argument clearly shows this is uncountable. The reason the rationals for instance are countable is by a map to their fractional representation. Attempting to map to the decimal representation would be fruitless.
Like 0.202020202...? Good point
I still don't get the conclusion why the new number cannot have already been part of this list
to be honest... the video has raised more questions in my mind than answered answer. lovely.
Like is she single
Wow, it seems you're making videos about every concept I've studied this year in my maths grade, and that's amazing! Great video!
Wow. Really great video. I had always known how decimals “worked”, but the zooming in analogy makes it so clear. And what’s also great about this is it shows how having the ability to think in other bases can be really useful. Thank you for expanding my understanding of this topic!!!
You should also mention that the Cantor function is not the only important function which is given the moniker "the Devil's Staircase". But they all have similar properties and it is a great example of such a function. It is probably the most famous version (possibly rivaled by only the Minkowski Question Mark function).
No discussion on the differentiability of the cantor function? I am somewhat disappointed :(
Jack Lam, agreed.
Well, it's not differentiable. At the points that make up the Cantor set, the slope becomes infinite.
The videi made it seem like it is differentiable almost everywhere. It is Lipschitz continous?
if you start with a function of slope 1 and flatten it in some areas then the maximum slope should stay one. The derivative is 0 at all points that are not in the cantor set, and 1 at the ones which are. Another way to look at it is that an infinite slope would mean a jump in the function similar to a traditional staircase function. This is not the case for the cantor function since it is contiguous
Ah I was halfright. It is differentiable almost everywhere (wiki says it differentiable up to a uncountable subset of the cantor set) but it is not Lipschitz continous
I feel it's worth mentioning the concept of Fractal Dust. Any Julia-set constructed where the origin is not contained over infinite iterations produces a dust (as I have learned, I may be wrong on this one). The dust has exactly the properties of the Cantor set (Which is also known as the "Cantor dust"); uncountably big but exactly 0 in "length" -> something I intuitevly think of as dimension between 0 and 1 (The Cantor dust having a dimension of approx 0.631). Cool stuff!
Great episode, probably the best one yet. Keep up the good work.
nth
Cynthia, well, you're not wrong...
(n+1)th
oh god
edit : (n+2)th
θth
(1/n!)th.
(Yes, that says 1 over n factorial.)
ωth
now take it's derivative.
Its derivative does not exist
The derivative is 0 everywhere except on the Cantor set where it's undefined.
That's not interesting. Find the area under the curve instead.
Area is 1/2 because it's symmetric
I meant a function for the area under the curve.
Georg Cantor is one of my heroes. He was a very religious person, and he was convinced that math was 'God's language'. Georg was fascinated by infinities. Even a cursory glance by anyone (you don't have to be math whizz!) shows that all the natural numbers 1,2,3,4 ... and all the even numbers 2,4,6,8 ... are both infinite series. (BUT is one twice the size of the other? :0) Cantor devoted his life to infinities and almost incidentally created Set Theory (or discovered it?); but his mental health suffered - and also his colleagues shunned him on religious grounds ("no-one knows the mind of God") rather than on his mathematical work. He died in an asylum unloved. There was a great vid of his biography on YT around - have a look or check out Wiki.
Yes, the Cantor function does have vertical leaps. They are in at infinity, but not on a finite amount of stages. It's like 1+1/2+1/4+1/8+... is equal to 2 when you do it infinitely many times, but not after a final amount of steps.
4:19 not true, a point can have a 1 in the expansion at the end of the expansion. just not anywhere other than the end. for example, 0.0220200020221000 is in the cantor set, whereas 0.0220200020221002 is not.
I noticed this too, but the explanation seems to be that every number whose ternary expansion ends on a 1 can be rewritten as ending on an infinite string of 2s, so 0.1 = 0.0222..., for the same reason that 0.999... = 1.
But you could also have 0.022020002022022222222222222222(22222)....., and that would be the same as 0.022020020221.
Ah, so the Cantor set is all the functions that have a base-3 representation without 1's, although some numbers will have a second base-3 representation with 1's. Or you could eliminate these points* by defining the Cantor set with open instead of closed intervals.
*edited to add
The Cantor set is all numbers in [0,1] which *can* be represented in base 3 using only 0 and 2 (some numbers have more than one representation in base 3, and we only need one of them to contain no 1s). So 0.1=0.02222... is in the Cantor set, and same with every other decimal which ends in a 1. But 0.11111... or 0.12121... would not be. (Defining the Cantor set with open instead of closed intervals would not work - you'd end up with a different set.)
Also, the Cantor set is what you get after repeating to infinity. So where is this one? At the end? What end?
Saying that cantor set has length 0 is like saying a line has 0 area. The cantor set is a fractal with dimension between 0 and 1. So it has infinite points but 0 length.
Yep. Cantor set is not weird if you learn about fractals
Shriram maiya, precisely. One of the best explain's I have ever heard.
To prove that the Cantor's Set is uncountable, I think you have to use the Cantor's Diagonal. First, you list all the numbers you think are in the set in base 3. Therefore, you'll have only numbers with 0's and 2's. Then you pick the first decimal (or "threemal" kkk) place of the first number and change it (if it's 0 you change to 2 and vice versa). You move to the next number and see the second decimal place, changing the values, and so on. After this infinite process, you end with a new number that is in the Cantor's Set and you haven't written down. So, the set is uncountable.
Oh, and I have to say: the videos are great and every time I watch a new video, more I like the channel!! It's very nice!!
(And sorry for my English kk I'm from Brazil)
Such a great channel and I have no idea how come I haven’t heard about it for so long.
4:15 _All the points with no 1's._
*Mind: blown*
At 4:14, you said the set contains all numbers with no 1 in the base three expansion. Shouldn't any number in the interval with a 1 only at the end of the number be in the cantor set as well, because it is the end point of an interval and therefore won't get removed? For example, although .1 has a 1, it will remain in the cantor set because it will be the endpoint of an interval at any stage greater than 0.
You can rewrite 0.1 as 0.022222222222...
Microwave44 Yep, exactly as stated above by Jonathan Gross.
Just like a number can have different expansions in base 10 (e.g. 0.5 = 0.50000... = 0.49999...),
there can be different representations for the same number in base 3 as well - just like 0.1 = 0.10000... = 0.02222... (in base 3).
So to be precise:
A point x between 0 and 1 is in the Cantor Set
if and only if
it has at least one expansion in base 3 where there are no 1's.
@Johnathan Gross
However it's an unstandard way of writing numbers, for me Kesley's statement is false.
It is very standard. Just because you didn't do it that way in elementary school doesn't mean you are right.
Just because it's valid, it does not mean it is "standard" or common amongst untrained mathematicians (which this show is aimed at). Also the graphics still show "0.1" and "0.01" etc and Kelsey says to remove the open interval, thus explicitly leaving the once with the 1 in the set. So it could have at least been mentioned in the video. I also tripped over that.
These videos are amazing! I've not had the chance to study cantor sets and fractal geometry in detail yet, but I am already very hyped. Am I right in thinking Cantor sets are clopen too?
Wonderful video, so clear and interesting! thank you!
...the autogenerated subtitles for this video are in Portugese.
0:02appreciate that by edicts which is a 0:04 Marcao master share were coming 0:08now I have to form those arriving is 0:19It had to be Latin for instance 0:30I had to go in 2011 12 0:49since 2000 and I will not give my 1:02better and had my guarana powder 1:14Children's inter milan in November
4:31, fractal and self similar are not interchangeable.
I know, right? 3blue1brown taught me that :D
Congrats for 100,000 subscribers!
Thank you for the awesome shout out! We hope your viewers like our channel if they trek over there to check it out. We are huge fans of Infinite Series -- and look forward to collabs in the future!
To prove the uncountability of the Cantor set, we can use Cantor's diagonal argument (a little bit ironic isn't it ?).
Let's imagine that the Cantor set is countable and that we can write all the numbers in the set in base 3. For example, we have :
0.2002222020202...
0.2222222222222...
0.0000200222222...
0.2222000000002...
.
.
.
0.0000000000002...
Now, for each n-th word of the set we change the n-th decimal with the other possible number (0 ->2 and 2->0) to create a new number. For our example, we get :
0.[0]002222020202...
0.2[0]22222222222...
0.00[2]0200222222...
0.222[0]000000002
.
.
.
0.000000...000000[0]...
This means that if we say that the new number is composed of the new values we got by changing every decimal of every of the number in the set, the new number would not be in the set, because we changed it by one value for each of the numbers in the set, despite the fact that it has all the properties needed to be in the set which in turn means that the set is uncountable because we have proven that we cannot construct a finite set which contains all the numbers in Cantor's set.
I hope you will see my proof and I'm sorry if I have done any errors in English it's not my first language. :)
Good proof but your last paragraph could be stated more clearly. If the new number is on the list then it must equal the n'th entry for some natural number n. But by construction the new number must be different from the n'th entry in the n'th digit, a contradiction. Therefore the number is not on the list.
Tinawyn658 Yes, but please learn what irony is
To prove the countability of the Cantor set, we can find a way to count the numbers in the Cantor set without missing any of them. As popular examples, we have the natural and - via some smart method - the rational numbers.
So here it is: The beginning of an infinitely long list that would contain _all_ numbers in the Cantor set (in base 3):
#1: 0
#2: 1
#3: 0.1
#4: 0.2
#5: 0.01
#6: 0.02
#7: 0.21
#8: 0.22
#9: 0.001
#10: 0.002
...
Now continue by showing that there can't be a number in the Cantor set that isn't on this list, and we're done.
The fact that cantor function has a length of 2 is so cool, since it is the as the length of a hypotenuse of a right triangle if the if we only allowing a a horizontal and vertical steps but not a diagonal step.
i.e. if we draw a right triangle ABC (90 degree on B)on a grid, where the hypotenuse is drawn as a stair, then moving from a vertix A to a vertix C on the hypotenuse will have the same length as moving from A to B then to C.(where only horizontal and vertical steps are allowed).
That's not a hypotenuse.
yeah, but it(the Cantor Function) has the same length as the triangle I'm talking about.
It's not a triangle. It has infinitely many sides, not three.
Exactly what I was thinking. It's like an infinite digital representation of a hypotenuse. Only a really weird one.
When only vertical and horizontal steps are allowed, it's called "continuous taxicab geometry" (en.wikipedia.org/wiki/Taxicab_geometry). It has some interesting properties, like that "circles" become squares rotated by 45°.
"Above the noise" sounds excellent. As long as it's not partisan.
Interesting remark about being big and small: there is also a topological version of being small: being the countable sum of nowhere dense set (it is called first Baire category set). Using similar construction but removing smaller and smaller bits of intervals in each step one can produce for any epsilon>0 nowhere dense set (homeomorphic with Cantor set) of measure 1-epsilon. Taking epsilon_n=1/n and summing over all n's yields an example of the set of full measure (being big in the sens of measure theory) which is first category (being small topological). You can repeat this construction with every interval [n,n+1] and therefore express the whole real line as the sum of two SMALL sets: one being measure zero, second being first catgeory.
There are numbers in the Cantor set with a 1 in their base 3 expansion -exactly those numbers with only one 1 in it, where the 1 is at the end.
For example 1/3 (0.1 in base 3) or 7/9 (0.21 in base 3). Intuitively (geometrically) those are the numbers exactly on the right edge after making a cut.
Patrick Wienhöft However, those numbers can also be interpreted as ending with an infinite number of 2's, similar to how 0.999... = 1.
Half-Pixel - by that argument there is no '2' in the real numbers cause you can write it as 0.999999 - it would also mean that the real numbers are not only not continuous but also countable infinite - which they are certainly not.
That's not at all what I'm saying. I'm saying that the base 3 expansion (or base anything expansion) isn't unique for numbers with a finite number of digits, e.g. in base 3, 0.22222... = 1. So all the numbers in the Cantor set have a base-3 representation with no 1's, even if that's not the only way to represent it.
That is not what the video was saying.
Kelsey specifically said that the cantor set includes no numbers containing a 1 in base 3 - and that statement is wrong as 0.1 base 3 is included. giving it a different expansion does not change that the numbers is still there - again - it would be like saying the real numbers do not contain the number '1' just cause you can write it in a different way.
You're right, she should have phrased that statement more carefully.
Yep, I understood some of the words...
All I came away with is Devils Haircut stuck in my head.
hahahaha just made my day
Wow... It's blowing my mind
Challenge question: All numbers in the cantor set can be written as a string of 0's and 2's.
1) List every number in the cantor set 0.0000000 to 1.000000....
2) Construct a number, n which is defined as follows: look at the third's place of the first number in our list. Our number will have a 2 if this is a 0 in it's third's place, and a 0 if this is a 2 in it's third's place. For the ninth's place of our number, check the ninths place of the second number on our list. If it is a 0, our number's ninth's place will be a 2, and if it's a 2 our ninth's place will be 0. More generally, for our number n, if we look at the m-th number on list at the (1/3)^m's place, n will have a 2 where it has a 0 and a 0 where it has a 2 at it's (1/3)^m's place.
3) N differs from every number on our list at some place. It differs from the 1st number at it's third's place, the 2nd at it's ninth's, and the m-th number at it's (1/3)^m's place. Therefore, it is not on our list.
4) Because this construction works for every arbitrary list, that means that all numbers in the Cantor Set cannot be listed. This means it is unlistable.
5) Therefore, the cantor set is uncountable infinite.
Considers just showing you could make a bijection to the binary representation of the real numbers, but I think proving it on it's own terms and self contained is more satisfying : )
4:18 "whose base 3 expansion contains no 1's", but of course there are examples onscreen that contain 1's. You might add the qualifier that they are allowed to have 1 only as their last digit. Alternatively, you might say that terminal 1's are rewritten as an endless tail of 2's, so for instance the point 1 gets written as 0.2222...
agree, the set should be: [0, 0.1> + [0.2, 1>
Well, it's still a closed interval. 0.222... is just a different way of writing 1 in ternary, it's not a different number.
If this Function is continuos, could you take its derivative? If yes , what would it be on the cantor Numbers?
Someone correct me if I'm wrong, but I think this function is continuous, but not everywhere differentiable. It is important to know that on a closed and bounded set, if a function is differentiable, it is continuous, and if it is continuous, then it is integrable. Keeping that in mind, I believe the derivative is 0 while on the compliment, and undefined everywhere else.
Note that while differentiability implies continuity, the converse is not true. In fact there exist functions which are everywhere continuous but nowhere differentiable.
The devil staircase is a BV function and thus its derivative, that can be defined as a distribution, is a measure: in this case the derivative on the cantor is the natural hausdorff measure of dimension log2/log 3. Outside the cantor it is 0
At each stage of construction of the Staircase, each diagonal piece has the same slope. That slope is therefore the total rise (1) over the total run of those pieces (which is the measure of the Cantor set). As the video states, that measure is 0. Thus rise over run is 1/0: that's a jump.
I would like to make two comments. The first is on the statement "All points whose base 3 expansion contains no 1's" (@4:15), and the second is on the subtractive way the set is constructed.
Clearly 1, 1/3 and 1/9 are points in the Cantor set, and their base 3 representation are 1, 0.1 and 0.01 respectively.
In fact both end points of every intervals in each stage included in the set, and all upper end points of the intervals ends with the digit 1 (in base 3).
At stage 0 the interval is [0, 1].
At stage 1 the intervals (written in base 3) are [0, 0.1] and [0.2, 1].
At stage 2 the intervals (again in base 3) are [0, 0.01], [0.02, 0.1], [0.2, 0.21] and [0.22, 1].
And so on.
So is the statement wrong?
Well, points with a finite base 3 expansion (actually, such expansion is also infinite, it just ends with infinite sequence of 0's) actually has 2 such expansions 0.10000... is also 0.02222222... (and I think it was mentioned (for base 10) in some previous episode).
So the statement is true when taking the infinite expansion (the one with infinitely many non zero digits).
Another remarkable property of the set and its construction is that it is not equals to the set of all end points (i.e. the infinite union of all end points of each stage) because such a union will only contains countable many points
Could you do a video about how the distance between Primes relates to Quantum Theory. What's going on there?
Red Yellow Ledbetter II 😲😲😲😲
A brief introduction to the question for the interested:
The relationship is due to renormalization and its relationship to the Riemann Hypothesis. Euler showed the relationship between the zeta function and prime numbers (more specifically their distribution). During renormalization in quantum theory, the same kind of infinite summations and analytical continuations come in to play. Extensive work has been done in renormalization theory to strengthen the assumptions. But Riemann's Hypothesis remains unproven (which is about the zeroes of the zeta function and those are really very important for the distribution of prime numbers)....
Sorry to be a Can short of a six pack, but what does this mean for QM. We need to see if the Riemann's Hypothesis is proven to see if it tells us any thing new about Quantum Theory?
I'd like to know that too! I only know the mathematics-- since I'm not a physicist I don't really understand what it means for the interpretation of QM. Basically it's an attempt to link fundamental numbers in quantum theory to the zeta function, motivated by the striking similarities in the mathematics and specifically unbounded integration (with weird infinite summations to finite values and all). So if the fundamental structure of QM really is the zeta function, then the question becomes what are the zeroes of the zeta function in QM? By RH and Euler these say something about prime numbers, so at least one vague hope is that primes could be a first principle from which to derive quantum numbers (at least in some sense). But beyond those rudiments, I don't know at all what that really means for the interpretation!
PS. I'm being a bit overly vague in the above; the zeroes are taken to be the eigenvalues of Hamiltonians in chaotic systems-- the latter being my area of study as a computer scientist dealing with stochastic processes in complex adaptive systems. If these can be shown to have a complex map to primes, then we can do some very interesting things in terms of simulation (although my intuition is immediately that this map will be NP-hard to compute).
4:15, should be no 1s in middle, meaning included only if ending with 1.
Isn't 0.1 included? because you said around 0:50 to remove open intervals!
brilliant presentation.
Treating the jumps in the staircase as vertical lines makes understanding it easy, the slope at the jumps approaches infinity
You keep repeating "almost everywhere" and at "almost every point", but you never mention that this phrasing in not just mathematical hand waving, but actually well defined mathematical terms (sure they require some measure theory to understand, but you did mention measure theory (at least indirectly) some times).
So Cantor's Set is like a one-dimensional fractal? Interesting.
Also, you could say that no one lives in it (Get it? No 1?)..... Sorry. :)
It's more like a zero-dimensional fractal.
I've seen some of the proofs on the uncountability of the Cantor set and while they make some sense they also confuse me. The Cantor set looks countable to me, here's why:
At each step let's look at the end points, we know from the video that the length of the set equals 0, which I interpret as all middle segments eventually get removed. Therefore the numbers in those segments are not part of the Cantor set (except the ones that are end points in a future step)
After 1 step we have the points: 0/3, 1/3, 2/3, and 3/3.
After 2 steps we have the points: 0/9, 1/9, 2/9, 3/9, 6/9, 7/9, 8/9, and 9/9.
Basically after n steps we have some subset of the numbers: i/(3^n) where i is between 0 and 3^n. These numbers are all rational.
To me this looks like all of the numbers in the Cantor set are rational and I know from various other proofs that the rational numbers are countably infinite.
Can someone explain what's wrong with that logic? At what point do irrational numbers get introduced? (As end points since the segments are always removed)
Just because every element of a sequence has some property does not mean that the limit of that sequence must have that property.
You're correct that at every step N, all new endpoints are rational, but that does not mean that the limit set must contain only rationals. The base-3 number 0.2202002000200002..., for instance, is irrational and belongs to the Cantor set.
In other words, all the irrational numbers are added on the "infinityth" step.
Proving it has length of 2:
Let x be the length of the Cantor function
Divide into 3 regions, (0,1/3), (1/3,2/3), (2/3,1)
Since the Cantor function is self-similar on domain (0,1), (0,1/3) and (2/3,1), we can write an equation like this:
x = (1/3)x + 2/3 + (1/3)x
Which gives x = 2.
isnt this just a 1d menger sponge?
SmashMan technichally 0.63D
When I see Real Analysis, sometimes I feel am I mad or the common people in the world are mad, or all of us actually living in an illusion or Maya 😮.
You should try non-standard analysis Ψ(｀▽´)Ψ.
this was awesome
Really enjoying the channel.
3 morons who understood nothing disliked this video. Great video by the way!
I agree they shouldn't have disliked the video, but this comment section isn't a place for condescension.
Qikun Xiang why did you assume the only reason they didn't like it is because they didn't understand it?
My bet is that at least one of them wanted to see a literal devil's staircase :)
Vadim Marushevskiy My bet is that at least one of them has a PhD in math.
AMAZING
My favorite thing about the Cantor set is that its points are also those that do not get "lifted" along a Koch curve. Go fractals!
I miss this show.
the set [0,1] that we start with is uncountable and is the same size as the real line, at each step we remove a finite number of smaller and smaller intervals. In the end , a COUNTABLE union of intervals of smaller size (decreasing monotone) is removed from an uncountable set, this will still leave us an uncountable set, akin to removing the set of natural numbers from (0,1) .Hence , the resulting Cantor Set is also uncountable.
Uncountablity of the cantor set:
Let associate to each integer a unique number of the cantor set like this:
1 -> 0.02222... this last number being written in base 3 (no 1 so it belongs to cantor set)
2 -> 0.20222...
3 -> 0.22022...
.
.
.
n -> 0.222...0...222... the "0" being at the nth decimal after the point
etc
All the integers have a unique number of the cantor set associated, however this number 0,220220222... (in base 3) for example also belongs to cantor set but has no corresponding integer
Hence the result (sorry for eventual bad English, not my native language ^^)
Nice vid, as usual. If Cantor function is continuous and has no lip, does it mean it is differentiable everywhere on the unit interval?
"All points whose base 3 expansion contains no 1s". I believe there are several exceptions: since all the slices of existence are inclusive, every number that is zeroes and then 1 (including 1 itself) ARE in this set.
I’m curious to the history of this function. This feels like a response to a larger idea or problem
To prove the uncountability of Cantor's set, after infinitely many divisions pair the points of each line segment with a natural number. You can then divide each new segment infinitely many times as well etc.
Musings: Suppose we also removed the endpoints of the intervals at each stage of the construction and called the limit the Tancor set. Is the Tancor set open, closed, or neither?
An "interior point" of a set is a point that has an open neighborhood entirely in the set. Note that no point outside the set can be an interior point but not all points in the set need be interior points, for instance 0 is not an interior point of [0, 1). The interior of a set is the union of all of its interior points, and a set is open iff it is equal to its interior. Like the Cantor set the Tancor set has an empty interior.
A "closure point" of a set is a point for which any open neighborhood intersects the set. Note that the closure point itself need not be in the set, for instance 1 is a closure point of [0, 1). The closure of a set is the union of all of its closure points, and a set is closed iff it is equal to its closure. The Tancor set is not closed because its closure is the Cantor set.
So the Tancor set has empty interior but isn't closed, which seems weird at first. But after some thought it's easy to come up with other such sets, for instance {1/n} for all natural numbers n.
great idea, good explanation. I appreciate your work/
Love you guyrls!
For a very pedantic nitpick, you need to say that the Cantor set consists of "real numbers which have a base-3 representation in which the digit 1 does not occur". To be clear why this nitpick, the number 0.1(base3) _is_ in the cantor set, because it has the representation 0.022222...(base 3). You do not get quite the same set if you remove _all_ reals whose base-3 representation contains a 1 somewhere (you lose the right hand point of every interval if you do this).
it felt weird until I realized that the number of points where the staircase rises is uncountably infinite. It's kid of like if Kronecker delta and horizontal line had a child after crazy crazy orgies...
there is no better way to start the weekend than to have Kelsey turn my brain to deep fried mush. 😨
I absolutely love stuff like this. Do these sorts of paradoxes - where looking at certain properties in one way gives completely incompatible answers when looked at in a different way - ever crop up in application? Or do they tend to just get filed under "huh, that's wierd" and then moved on from?
Gabe Tower it's not a paradox, it just seems odd because math doesn't work the way we might be expect it to
paradox
noun
a seemingly absurd or self-contradictory statement or proposition that when investigated or explained may prove to be well founded or true.
It is in fact a paradox.
It's actually not too hard to understand using basic function rules. It has length 2 because it's actually the same as the slope that goes horizontally then vertically, just broken up a lot. The across parts are all the flat inbetween lines, and all the canter set points are direct vertical lines. Since these vertical lines only happen at infinitely many *single* points, it still passes the vertical line test and is thus a function.
wow! You've really got alot of subscripers since the start. Felt like yesterday this channel started. Good work!
All the numbers in the Cantor set (in base 3) have the numbers 0 and 2. Change every 2 to a 1 and you have all the numbers from 0 to 1 (in base 2).
Those two sets are bijective because:
a) no binary number can be reached twice (trivial and inconsequential)
b) every binary number is reached at least once, because for every binary number from 0 to 1 the number (in base 3) that has a "2" instead of the "1" is part of the cantor set.
By definition, the numbers that belong to Cantor’s set Κ are uncountable if we can find a one to one correlation with an uncountable set of numbers, such as the interval of real numbers between 1 and 0.
Our correlation is the next:
Since the decimal expansion of all Cantor set numbers contains no 1, we are left with all combinations of 0 and 2. By substituting the 2 with a 1 we have obtained a one to one relation between the Κ and all base 2 real numbers I. In fact, we have substituted all combinations of 0 and 2 with all combinations of 0 and 1.
This proves our theorem.
I think an easy way to prove that the Cantor set has any points at all is to look at the endpoints. They're never removed, so they must always be there. This also includes base-3 numbers like 0.1, 0.01, 0.21, 0.001, 0.021, 0.201, 0.221 etc. If the last digit is 1, it is allowed as well.
My inability to argue coherently against "zeroeth" doesn't imply that I am resigned to accept it.
Just one thing, to be more precise: in the Cantor set we do not have numbers with 1 in their base-3 expansion because we can replace every one in the endpoints of some of its intervals with 0.2222222... so that no 1 is present.
The uncountability of the Cantor set can be proved with diagonalization, as the others have already shown.
P.S. fractals are not only about self-similarity. they're more about the roughness of a shape even at the infinitesimal level. :) I wonder if it could be made an episode about this...
In other words: 0.999... = 1 but for base 3.
Exactly :)
Simple proof for challenge question :
0 to 1 contains uncountably infinite numbers. Now after we remove 1/3rd elements from middle two new line segments are formed but these are also uncountable infinite . In fact no matter how small the distance between 2 points on real line is, it will always contain uncountably infinite numbers. Thus no matter how many times we remove those 1/3rd pieces from number line, we will always have uncountably infinite numbers.
There are so many funny thinks with the continuum that you have to ask:
1. Is it possible that the continuum idea is wrong? maybe there is some kind of Russell's paradox that we have not seen. Not so much because of the Cantor Function but because of the The Banach Tarski "paradox"
2. It's pretty clear that reality is unlikely to be based, or isomorphic to any model based on continues functions.
4:16 it's only here that it's clear you were removing closed intervals from the original closed intervals (or at least closed on the left). If you were removing open intervals then you would still have base 3 numbers with 1s, namely those that have expansion that ends with a 1.
Have you/would you do a video on Julia sets, or the Mandelbrot set? I'm curious to learn more about it. Or something about fractals in general?
Funny enough, you can proof that the Cantor set is uncountable by Cantor's diagonal argument.
Suppose the set was countable. Then we could match each positive natural number with a number in the Cantor set, e.g.
1 -> 0.2020200202220...
2 -> 0.220002020202...
3 -> 0.002200200202...
4 -> 0.022002200220...
etc
Now take the first decimal digit from the first number, the 2nd from the 2nd number etc. and put 0. in front of it
For my example: 0.2220...
Now swap all decimal digits: 0.0002
The number gotten by doing this is different from any number on the list, so it's not on the list. Therefore our supposed matching of natural numbers and the Cantor set is not a bijection, therefore they have different sizes (Cantor set is bigger), therefore the Cantor set in uncountable.
This series tends to give example sets in threes or fives which, personally, I would love to see increased to fives and sevens or similar just to get a better feel of [cantor] sets, number series, etc.
Good one
You can define an isomorphism between the reals in [0,1] and the Cantor set by replacing every one in the binary expansion with twos in base 3. So they have equal cardinality.
christ I wish you were my math teacher in high school... I needed all my brainpower just to process the meaning of the words you were using (not a native... I am a translator, BUT math and science is not my field of competence), but it was very interesting!
At first the length of the function confused me. My gut feeling (which apparently is adjusted to bizar math) was that the length would be 1. So it surprised me at first when you said it was 2. But then it dawned on me that the sloped parts must have an infinite slope, just as that length 2 line.
Thanks for the video
'Splaining the Devil's Staircase to non-mathematicians?...that's ambitious. I first ran into this taking a course on fractals using Barnsley's "Fractal's Everywhere" as the text book. It leads one into some pretty deep stuff.
Simplest proof of unaccountably many points in the Cantor Set:
In each step of construction, the number of points in that step is uncountable. By induction, the Cantor Set is uncountable.
(This assumes that the reader is familiar with the interval [0,1] being uncountable)
This video is so cooooooooool!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
step 1: see proof of uncountability of real numbers between 0 and 1
step 2: proof that you can turn any decimal number into a number in the cantor set: a) pick a number x. b) write it in binary. c) replace every 1 with a 2. d) this number can't have a 1 in it, because it would get replaced by a 2, which means it's in the Cantor set.
step 4: since the above produces a different point in the Cantor set for every real number 0
The ternary expansion of the points on the cantor set only have zeros and twos. Now associate this with the binary number that's the same except the twos become 1s.
Eg: 0.00202020202 goes to 0.00101010101.
You can also do the reverse to any real binary number between 0 and 1.
Eg: 0.1010011 goes to 0.2020022.
We've already proved why there are an uncountable number of numbers between 0 and 1.
It is easy to construct a bijection from [0,1] into the Cantor set.
Given x in [0,1], consider x0, x1, x2, ... its digits in basis 2. Associate to x the number which has digits 2*x0, 2*x1, 2*x2, ...