The Devil's Staircase | Infinite Series

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Topology Riddles
• Topology Riddles | Inf...
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
• Topology Riddles | Inf...
Ralph Dratman
• Topology Riddles | Inf...

[@BothHands1]

Math is awesome when it's not mandatory xD

[@rafaelazo75]

Ye

[@cameron6803]

This is the realest thing

[@davidwright8432]

How true! Also, most math teaching is ... well, the moderator might ding me. Just let's say, highly student-hostile.

[@JamesFerguson72]

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.

[@GunsExplosivesnStuff]

Actually there's this really cool things that all humans are equipped with called memory!

[@-danR]

GunsExplosivesetc. Yeah, I've memorized all my contacts' phone numbers, addresses, birthdays, favorite sushi's and colors...

[@BobfromSydney]

In between the hundreds of clickbaity titles filling up our KZbin 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.

[@snbeast9545]

It's a strategy to get more views. The red bar that covers the place where KZbin'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.

[@aishaaofthedays]

@@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...

[@bockmaker]

What do you play at the top of the Devil's Staircase? Gabriel's Horn.

[@lucashoffses9019]

I actually play the horn, but my name isn't Gabriel.

[@harryandruschak2843]

For those who may not know, "Gabrieel's Horn" is a mathamatical construction with finite volume and infinite sueface area. See WIKIPEDIA

[@ilmbrk6570]

bockmaker good one

[@OskarElek]

you play Rocky's theme song

[@sciblastofficial9833]

Ilm Brk If you don't get it, Gabriel's Horn is a weird object also a fractal

[@atimholt]

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.

[@luiz00estilo]

Lol, I thought the proof was very complex. That's neat

[@SaveSoilSaveSoil]

Neat. A surjective map from Cantor to the unit interval.

[@tsawy6]

@@SaveSoilSaveSoil (actually pretty sure it's bijective)

[@ShehabEllithy]

@@tsawy6 She did not say it's *not* injective, so she did not say anything wrong.

[@tsawy6]

@@ShehabEllithy Sure, I wasn't tryna correct so much as expand

[@1JDRM]

The way she explained what a base is and how it works is outright brilliant.

[@enclave2k1]

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

[@CarmenLC]

nth

[@asthmen]

Cynthia, well, you're not wrong...

[@Euquila]

(n+1)th oh god edit : (n+2)th

[@chiyanyu553]

θth

[@ganaraminukshuk0]

(1/n!)th. (Yes, that says 1 over n factorial.)

[@phucminhnguyenle250]

ωth

[@binaryblade2]

now take it's derivative.

[@firefly618]

Its derivative does not exist

[@martinepstein9826]

The derivative is 0 everywhere except on the Cantor set where it's undefined.

[@bloodyadaku]

That's not interesting. Find the area under the curve instead.

[@firefly618]

Area is 1/2 because it's symmetric

[@bloodyadaku]

I meant a function for the area under the curve.

[@paradoxica424]

No discussion on the differentiability of the cantor function? I am somewhat disappointed :(

[@asthmen]

Jack Lam, agreed.

[@JohnathanGross]

Well, it's not differentiable. At the points that make up the Cantor set, the slope becomes infinite.

[@zairaner1489]

The videi made it seem like it is differentiable almost everywhere. It is Lipschitz continous?

[@Jonasjones100]

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

[@zairaner1489]

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

[@asthmen]

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.

[@MrMctastics]

Asthmen according to my browser anyway, yours is the first proof. Seems right to

[@asthmen]

[@Tumbolisu]

sort by "newest first", scroll down, yes its first.

[@wren1728]

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.

[@CurlBro15]

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.

[@TheRealTNorty]

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.

[@shvoregavim9435]

Nice proof!

[@BRUMARTUBE]

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!

[@Ronnypetson]

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.

[@7lllll]

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.

[@bobbobson2061]

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.

[@lucashoffses9019]

But you could also have 0.022020002022022222222222222222(22222)....., and that would be the same as 0.022020020221.

[@martinepstein9826]

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

[@elizabeth8720]

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.)

[@tomc.5704]

Also, the Cantor set is what you get after repeating to infinity. So where is this one? At the end? What end?

[@ThomasOrlita]

Yep, I understood some of the words...

[@brokenacoustic]

All I came away with is Devils Haircut stuck in my head.

[@kingdick2855]

hahahaha just made my day

[@TheLoneGnu]

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).

[@Ecrowtist]

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.

[@JohnathanGross]

You can rewrite 0.1 as 0.022222222222...

[@SillyTalent2]

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.

[@nUrnxvmhTEuU]

@Johnathan Gross However it's an unstandard way of writing numbers, for me Kesley's statement is false.

[@JohnathanGross]

It is very standard. Just because you didn't do it that way in elementary school doesn't mean you are right.

[@tetsi0815]

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.

[@acorn1014]

4:31, fractal and self similar are not interchangeable.

[@florianwicher]

I know, right? 3blue1brown taught me that :D

[@AndrewGordonBellPerc]

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.

[@iwersonsch5131]

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.

[@ClaudioGrecoPhD]

Did you just use the Cantor diagonal argument over the Cantor set? Cantor-ception!

[@huckthatdish]

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.

[@iwersonsch5131]

Like 0.202020202...? Good point

[@iwersonsch5131]

I still don't get the conclusion why the new number cannot have already been part of this list

[@kyzer42]

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. :)

[@Zero-es-natural]

It's more like a zero-dimensional fractal.

[@Tinawyn658]

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. :)

[@martinepstein9826]

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.

[@vernement4752]

Tinawyn658 Yes, but please learn what irony is

[@iwersonsch5131]

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.

[@kadourimdou43]

Could you do a video about how the distance between Primes relates to Quantum Theory. What's going on there?

[@TykoBrian7]

Red Yellow Ledbetter II 😲😲😲😲

[@ontoverse]

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)....

[@kadourimdou43]

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?

[@ontoverse]

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).

[@user-iu1xg6jv6e]

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!

[@shrirammaiya4381]

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.

[@shrirammaiya4381]

Yep. Cantor set is not weird if you learn about fractals

[@shrirammaiya9867]

Shriram maiya, precisely. One of the best explain's I have ever heard.

[@TheDavidlloydjones]

No problem. Fairies have wings and dragons breathe fire. And neither one of them exists. What's all the fuss about? Even the mathematics of it is simple. One, fairies. Another one, dragons. Both plural. Total number of both: zero. OK?

[@curtiswfranks]

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).

[@glutinousmaximus]

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.

[@epicsmashman6806]

isnt this just a 1d menger sponge?

[@joelhaggis5054]

SmashMan technichally 0.63D

[@jaspreetsingh-nr6gr]

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.

[@PlayTheMind]

4:15 _All the points with no 1's._ *Mind: blown*

[@HrishikeshKulkarni]

Would we see the same behaviour by taking out the nth part from the middle of each interval and looking at it from base-n? Where 'n' is odd of course. What I'm trying to ask is, can Cantor's set be generalised for any odd number which in this case was three?

[@vedantsinha7374]

to be honest... the video has raised more questions in my mind than answered answer. lovely.

[@13371138]

Like is she single

[@martinepstein9826]

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.

[@Khazam1992]

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).

[@JohnathanGross]

That's not a hypotenuse.

[@Khazam1992]

yeah, but it(the Cantor Function) has the same length as the triangle I'm talking about.

[@JohnathanGross]

It's not a triangle. It has infinitely many sides, not three.

[@wingracer1614]

Exactly what I was thinking. It's like an infinite digital representation of a hypotenuse. Only a really weird one.

[@xenontesla122]

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°.

[@Grato537]

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)

[@CM-lr7tf]

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.

[@patrickwienhoft7987]

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.

[@half_pixel]

Patrick Wienhöft However, those numbers can also be interpreted as ending with an infinite number of 2's, similar to how 0.999... = 1.

[@ABaumstumpf]

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.

[@half_pixel]

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.

[@ABaumstumpf]

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.

[@half_pixel]

You're right, she should have phrased that statement more carefully.

[@touisbetterthanpi]

I’m curious to the history of this function. This feels like a response to a larger idea or problem

[@Tehom1]

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...

[@tabularasa0606]

agree, the set should be: [0, 0.1> + [0.2, 1>

[@Tehom1]

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.

[@dudaheyhey]

You forgot the first thing of cantor set is the point 1/3 is in the cantor set or not ? Since in base 3 it's 0.1 so she suppose to be out but she is still in cause I can present it as 0.0222222222... so important:) but great explanation!! Your super

[@subhoghosal7]

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 😮.

[@derekmaggard9235]

You should try non-standard analysis Ψ(｀▽´)Ψ.

[@a2rhombus2]

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.

[@Shyguyyyyy]

3 morons who understood nothing disliked this video. Great video by the way!

[@MrMctastics]

I agree they shouldn't have disliked the video, but this comment section isn't a place for condescension.

[@mohamedlaminebouaziz5969]

Qikun Xiang why did you assume the only reason they didn't like it is because they didn't understand it?

[@VadimMarushevskiy]

My bet is that at least one of them wanted to see a literal devil's staircase :)

[@mohamedlaminebouaziz5969]

Vadim Marushevskiy My bet is that at least one of them has a PhD in math.

[@MalcolmAkner]

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!

[@KohuGaly]

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...

[@nbrader]

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.

[@rkpetry]

[08:09] plus-"we're left with all-these" infinitesmally-tall, infinite-biggly-sloped, steps, ...nothing weird about zero times infinity equals zero (zero is-not a natural number)... [09:51] no-it is-not, a continuous, function-it has infinitely many steps, not slopes... *[They're logically different, infinitesimal rise over infinitesimal run, vs. over zero run]*

[@rkpetry]

Maybe we should bring back that first-year-calculus worry, ∫ dx² (0,+∞) ≈ 1 .

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.

[@rkpetry]

[04:09] 'hungh'-these are closed, subsets → they include one '1', not no '1's... [11:33] probably not-until you find 'white coffee' (not of overcooked beans)...

[@rkpetry]

Try simplifying Cantor's infinitesimal ramparts by using n slides instead of 2ⁿ.... You could also try cosine'd slopes, instead of Cantor's sharp-edge uniform-slopes, that'd be continuous and derivative continuous everywhere....

[@saamspam6127]

Isn't 0.1 in base 3 in the cantor set even though it has a 1 in it?

[@jursamaj]

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.

[@ralphinoful]

I really dislike the proofs I saw in the comments, so I want to try and prove the uncountability of the cantor set using a Topological argument. Proof: Assume for the sake of a contradiction that the Cantor set is countable. This implies that there exists a bijection between the cantor set and the natural numbers. Let the cantor set be denoted as the set C, and the naturals N. So there exists some bijective function f : C --> N. Likewise, a bijective function g : N --> C. Consider the bijective function g. I will now prove that g is continuous. Take any open set in C, call it U. Then the pre-image of U is some set V in N. But what do open sets in N look like? Open sets in N (under the standard subspace topology) are just singletons. So V is just a collection of singletons, and any arbitrary union of open sets is open, so V must be open. So this means that the pre-image of our arbitrary open set U is open in N. So g is continuous. Since g is continuous, and a bijection, we get a contradiction. N is not compact (This is obvious and I won't prove it), and C is compact. A very easy proof for the compactness of the cantor set is as follows. The cantor set is closed because it's compliment is just a union of arbitrarily many open sets, and thus open. So since it's compliment is open, the cantor set is closed. It is a subset of R, and is bounded, obviously. So the cantor set is a closed and bounded set in R, and thus by the Heine Borel Theorem, it is compact. So g is a continuous bijection from a non-compact set to a compact set. Contradiction. Our assumption "The Cantor set is countable" leads to a contradiction, so it must be false. So the Cantor set is uncountable.

[@prateekgargx]

I can't stress how important this series is. Execution is brilliant and I wish more people can appreciate the content.

[@gJonii]

0.1(in ternary) belongs in cantor set, no? Edit: right, took me a bit, but cantor set only removes 1's that aren't followed by infinite number of 0's

[@Fetrovsky]

"Above the noise" sounds excellent. As long as it's not partisan.

[@BadBoy-ou3pw]

I like when people pronounce 0 as ZERO and not o O

[@NoNameAtAll2]

Cantor set isn't 0-dimensional (group of points), it is bigger Cantor set isn't 1-dimensional (a line), it is smaller Cantror set becomes 1/2 the amount if you zoom into 1/3 of it Cantor set is log3(2)-dimensional

[@jamez6398]

"It's a fractal, as in it is self-similar." Not all fractals are self-similar... Is the cantor set length 0, or is it infinitesimal? Those are not the same thing... Infinitesimal is 1/[infinite], instead. When x approaches infinity, then no approaches 0, but doesn't reach it. It's asymptotic, so it never reaches 0, as 1/[infinite]=/=0...

[@acarril]

A proof by contradiction: Let [0,1] be countable. This means we can list its elements as a sequence (x_n). Now lets pick a subinterval [a_1, b_1] of [0,1] with length less than 1/2 and that excludes x_1 (the first term of the sequence). Now we pick a subinterval of [a_1, b_1] and we call it [a_2, b_2], and we pick it in such way that it is less than 1/4 and excludes x_2. Continuing inductively, we can choose an interval [a_{n+1}, b_{n+1}] which is a subset of [a_n, b_n], has length less than 1/(2^{n+1}) and excludes x_{n+1}. We now have [a_1, b_1] which is a subset of [a_2, b_2] ... which is a subset of [a_n, b_n]. Since we have a decreasing sequence of nonempty closed sets with a diameter that tends to 0, this implies that the infinite intersection of these closed sets must be equal to a singleton. However, the infinite intersection of [a_n, b_n] is empty, which can't be, because [0,1] is complete. This is a contradiction, so [0,1] must be uncountable.

[@ricardofeliperosadacanesin4742]

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)

[@charlien.5841]

You can use Cantor's proof of why the irrational numbers are uncountable infinite to prove the uncountability of the cantor set. Just use the Cantor diagonal. Switch all zeros to twos and all twos to zeros.

[@johngrey5806]

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?

[@dudelookatree]

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.

[@inferno2046]

How can we possibly say that this function has no slope? How can we say that this function to the 10th is equal to zero if you can never actually get there

[@maimere]

I miss this show.

[@katakana1]

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.

[@fargoth_ur7]

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!

[@Wout12345]

Proof: All real numbers in between 0 and 1 have a binary representation consisting of only 0's and 1's. By replacing all the 1's with 2's and reinterpreting their values as ternary, all new numbers are now part of the Cantor set and all old numbers map to exactly one new number. Therefore, the Cantor set is at least as big as the real numbers in between 0 and 1, which can be shown to be uncountably infinite using Cantor's diagonalization argument (or you could directly apply it to the Cantor set too, but I like reusing results :P).

[@nathanisbored]

for the question: just use the diagonal argument again. make a list of base 3 numbers with only 0's and 2's as digits, and change all 2's to 0's and all 0's to 2's along the diagonal

[@MrJimmyd500]

FRACTALS ARE NOT NECESSARILY SELF-SIMILAR!!!!!! AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAARRRRRRRGGGGGGGGGGGGHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@Sk8aterBoy132]

Proof that the Cantor Set is uncountable: TL;DR non-Cantor Set numbers map to a countable subset of [0,1], namely, multiples of powers of 1/2. Therefore, the Cantor Set maps to an uncountable subset of [0,1], and must itself be uncountable. The Cantor Function is continuous. The range of the Cantor Function includes 0 and 1. (C(0) = 0; C(1) = 1) Therefore the range of the Cantor Function includes the closed interval from zero to 1. (For any y in [0,1], there exists an x, also in [0,1], such that C(x) = y) For any x NOT in the Cantor Set, C(x) can be expressed: C(x) = q/(2^n), where q and n are both natural numbers. (and q

[@alwinpriven2400]

assume the cantor set is countable, therefore you can make a bijective function F from the naturals to the set: F(0): F(0)(0) F(0)(1) F(0)(2) ... F(1): F(1)(0) F(1)(1) F(1)(2) ... . . . we will make a function G that does not appear in the range of F G = λn∈ℕ. 0 if F(n)(n)=2, 2 if F(n)(n) = 0 the function returns a number in the cantor set because the number does not contain a 1. we will now prove that ∀n∈ℕ. F(n)≠G Let n∈ℕ. If F(n)(n) = 0 G(n) = 2 ≠ F(n)(n) If F(n)(n) = 2 G(n) = 0 ≠ F(n)(n) We have found that for an arbitrary n F(n) ≠ G because they do not have the same value for n. therefore we have found a function that is not in the range of F and thus F is not bijective. Contradiction, therefore the cantor set in uncountable. ■

[@Danicker]

Wait, so in a Koch snowflake: each side is just part of cantor set, so the whole shape should have a perimeter of 0, but it also has an infinite perimeter because each iteration increases the perimeter by a factor of 4/3. Which is right?

[@warmpianist]

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.

[@sorokinmisha007]

Here's the proof: 1. Cantor's set contain all real numbers on [0, 1], whose ternary expansion contains no '1'. 2. For each number k in the Cantor's set, write out k's ternary expansion, replace all '2' with '1', interpret resulting string as a binary expansion (for example, 0.0202002000... base 3 becomes 0.0101001000... base 2). This operation is a bijection of Cantor's set and a closed interval [0, 1]. Therefore, Cantor's set and closed interval [0, 1] have the same cardinality (they contain the same number of elements). 3. tl;dr: Cantor's diagonal argument. Suppose we can count all numbers on [0, 1]. Then let L be a numbered list of all real numbers on [0, 1], L_k - k-th element on this list and D_k - k-th digit in a decimal expansion of L_k. Let real number R be defined by a decimal expansion 0.N_1 N_2 N_3... (more formally, R=\sum_{k=1}^{\infty} (N_k)/(10^k) ), where N_k - any decimal digit not equal to D_k (for example N_k = 1 if D_k = 0, otherwise N_k = 0). R is not on L, since it's different from any number on L (k-th digit of R is not equal to k-th digit of L_k by definition). That contradicts our assumption that all real numbers can be listed, therefore our assumption is false, and set of real numbers on [0, 1] is uncountable. 4. Since Cantor's set and [0, 1] have the same cardinality, Cantor's set must be uncountable. QED

[@razor1uk610]

..and there was I ...initially thinking this was about the Devil's Staircase Incident.

[@TheLoneGnu]

Solution to the challenge problem: Let C be the Cator set, and let c be an element of C. We saw in the episode that there exists an infinite series a_1,a_2,... in {0,2} such that c = a_1*3^(-1) + a_2*3^(-2) + ... (that is the ternary expansion of c). This expansion is unique (normally such an expansion would not be unique since for example 0*3^(-1) + 2*3^(-2) + 2*3^(-3) + 2*3^(-4) = 1*3^-1, the anternative form of an expansion would always contain the ternary digit 1, so in this case it is unique). Since this expansion is unique we can define the function f : C -> [0,1] as follows: f(a_1*3^(-1) + a_2*3^(-2) + ...) = a_1*2^(-1) + a_2*2^(-2) + ... Now recall from the episode that we also have that any number with such an expansion would be in the Cantor set. But that means that f maps the Cantor set (via the binary expansion) onto (surjectively (= hitting all numbers of)) the interval [0,1]. We have seen in an earlier episode that [0,1] is uncountably large. If C was countable, we could clearly not map it onto an uncountably large set, so C must be uncountable itself!

[@ddbhaha1529]

Simple. It's like asking how many numbers are there between 0 and 1. 0.2 0.22 0.222 ... 0.2......2 Just add a 2 at the end. Then it is a new number. So.... uncountable. You can also insert a 0 in the middle... for infinite times. Then it is also a new number. 0.2 0.02 0.002 .... 0.00000....2

[@danielbenyair300]

Really?!... you don't understand infinty do you... There are infinte points in wich the slope change so it DOES clime up in INFINETLY SMALL STEPS!!! example for infinty: if you try to draw infint lines inside an area you WILL end up filling the area!! so it wont look like lines AT ALL!!!!

[@damienperolle2138]

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 ^^)

[@dylanrambow2704]

I was literally reviewing the Cantor set just a couple days ago. Here's my crack at the proof that the Cantor set is uncountable, without referring to any books (this is great practice for my qualifying exam next month): Let {c_k}, as k ranges from 1 to infinity, be a countable listing of the Cantor set C. We will find a point x in C with x not equal to any of the c_k's, and this will contradict the listing of the set C, showing that it is uncountable. The point c_1 must lie either inside the interval [0,1/3] or the interval [2/3,1]. Pick x to lie in whichever interval c_1 does not. So, without loss of generality, we can assume c_1 is in [0,1/3] and then x is in [2/3,1]. Thus x is not equal to c_1. If any of the other c_k's lies in [0,1/3], then x is not equal to those c_k's either, and in this case we're done. So we can assume that c_k is in [2/3,1] for all k>1. Now c_2 is in either in [2/3,7/9] or [8/9,1]. Choose x to lie in whichever interval c_2 does not. So wolog assume c_2 is in [2/3,7/9] and x is in [8/9,1]. So x eq c_2. We can then, similar to what was done in the previous step, assume that c_k is in [8/9,1] for all k>2. Since each interval that we place x into in the kth step is then divided in the (k+1)th step into two disjoint intervals, we can continue this process ad infinitum. Whichever subinterval contains c_k in the kth division of C, we can choose x to be in a different, disjoint subinterval. This process will discover a point x in C that is not in the enumeration {c_k}, proving that C must be uncountable.

[@patrickwienhoft7987]

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.

[@estebanzd9434]

So, I was trying to do this in one sitting, but I couldn't. Let me show you. Since you are removing a third of every 2^n pieces, we have: ∞ Σ (2^n)/(3^(n+1)) = 1 n=0 let's call this sum S. So, 1 - S = 0 (our length). What about the slope? considering that 1-S is just numbers with no "x" or constant whatsoever, the derivative is 0. Is this a good response? I'm just in high school and I'm learning this for fun.

[@thegreyknight4282]

Just a question from a non-mathematician: In the case of Cantor's set, we begin with an interval. As far as I know, intervals always have infinite points in them, because it relates to a "From... , To..." form. (Infindecimals). And all that Cantor's set does, is with each step, dividing up the current interval(s) into two intervals with a sum of 2/3 to former interval(s). But, because you end up, never getting rid of intervals, just making them smaller, you always keep the property of "infinidecimality"(?). Now the big question: Is this correct? Is it a valid proof? If no, where's the flaw in my chain of thought? (I really have no idea ^^)

[@josugambee3701]

Let's call the Cantor Function can(n), with n being depth. So can(n) returns (can(n-1)/2, 1/2, can(n-1)/2) or something like that. I don't know if recursive functions are a thing in "proper" mathematics, though.

[@beirirangu]

just to be clear, are you saying Stage 1 is {0≥a1≥1/3 , 2/3≥a2≥1} or {0≥a1>1/3 , 2/3≥a2>1} ? because if the former is true, then it DOES include base 3 numbers that have the number 1 And there are several ways to answer the question, like, if you say there IS a highest number, then apply a further stage, it implies that there can't be a highest number, and the list in base 3 of all number possibilities, but I'm not as knowledgeable on that proof

[@lyrimetacurl0]

Interesting, it contains 2^n points which could imply that it has 2^(Aleph Null) points and that this is in fact not Aleph 2 (which is the number of real numbers between 0 and 1). Or it could be argued that it is the same because they both go from 0,0 to 1,1.

[@jonathanlimm7221]

Is it a function, though? The slope at any point in the Cantor set is infinite, right? Of course, for any specific point in the Cantor set, the distance that the function leaps vertically is infinitely small, so one does get to call it "continuous", but does the derivative showing singularities, both zero and infinity, not pose some kind of problem?

[@joshbasserabie6590]

Jonathan Limm it is not differentiable everywhere but it's still a function

[@JohnathanGross]

A function just means that each point in the domain corresponds to a single point in the range, the vertical line test. The slope may be infinite at the points that make up the Cantor set, but the value of the function is still well defined there. The value of both f(1/3) and f(2/3) is 1/2.

[@davejacob5208]

to me it seems that it is in fact a staircase, but it simply has infinitesimal small hights between its steps.

[@JohnathanGross]

Every function has infinitesimally small heights between it's infinitesimally small steps. That's how The definition of the derivative works. For each infinitesimal length dx, there is an infinitesimal rise dy.

[@jonathanlimm7221]

So, does that mean it is a non-analytic function?

[@t.e.d]

Proof that Cantor set is unaccountably infinite: Modify the Cantor set such that you remove eleven 12th's not two 3rds. Now after every step we are multiplying the number of points we have in the set by 11. Let's give every child of a point after a new step a name (0-9 and a). Now we can start naming all the points at discreet steps. At step 1 we only have a point named 0. At step 2 we have 00, 01, 02, ... 09, 0a. At step 3 we have 000, 001, ... 021, 022 ... 0a9, 0aa (cross product of all previous points with 0-9 and a). Since there are infinitely many steps in the set, we can assume we cannot match every number with a unique point since there are 11 points now for each 10 digits we have. For example in step 2 we are trying to match 0-9 with all unique points in step 2 and we get: 0 : 00 1 : 01 2 : 02 3 : 03 4 : 04 5 : 05 6 : 06 7 : 07 8 : 08 9 : 09 ? : 0a But we are left with 0a which cannot be counted with 1 digit, making the set uncountable. Continue this at any step other than 1 and you will find a number cannot be counted with n-1 digits where n is the number of steps in we are. The point of working in base 11 here is to use Cantor's diagonal theorem. It is still countably infinite in base 3 because by dividing countable infinity by the conversion from base 11 to base 3 is still countable infinity.

[@Zerotan]

Do all the crazy Cantor set properties hold if you choose a different sized chunk to cut? Like if instead of the middle 1/3, you cut out the middle 1/1,000,001 (for symmetry) or the middle 999,999/1,000,001? Are these still infinitely short, and uncountably infinite, and equally devilish in their staircase?

[@mattstokes3881]

Proof that the Cantor set is uncountable. Suppose we could list all the elements in the Cantor set (in their base three expansion). Then we can use Cantors diagonalizatoin argument (since we are in base three send 0 to 2, and 2 to 0 avoiding 1's). Example: Suppose this is our list of the Cantor set in base three. 1: .020202020... 2: .222220222... 3: .000002000... 4: .222220222... 5: .022022022... and so on Diagonalizaion gives us the element .20200.... which will differ from each element in the list in at least one place (and has no 1's). Thus this is an element of the Cantor set not on the list, and so the Cantor set is uncountable.

[@MaraK_dialmformara]

You can prove the uncountability of the Cantor set using the same list proof as for the set of real numbers. Imagine you could list all the numbers in the base-3 expansion of the Cantor set (i.e. all the points in the set). That list would look like this: 0.00000... 0.22002... 0.20220... 0.00200... 0.20002... You can construct a number that's not on the list but still in the Cantor set by making its first place different from the first place of the first number, the second place different from the second place of the second number, and so on. 0.20020... And because you can just keep generating numbers from the set that aren't on the list, the set is uncountable. PS this is like proving the uncountability of the set of real numbers using base 2

[@Bellezzasolo]

Write down the complete cantor set, assuming it is countable. Now, work diagonally down your list, taking the first ternary digit of the first number, the 2nd of the 2nd number, and so on. You now have something like this: 0.202220002022... If a digit after the ternary point is 2, replace it with 0. If it's 0, replace it with 2. You now have a cantor number that differs by a digit from all entries on your supposedly complete list... so the cantor set is uncountable. QED.

[@lucashoffses9019]

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.

[@Homar30]

Cantor Function is bijection of [0,1] interval and Cantor set. As [0,1] is uncountable, this proves Cantor set unaccountability.

[@ronenfishman510]

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

[@seanfraser3125]

My proof that the Cantor set is uncountable. This is more or less identical to Cantor's proof that the reals are uncountable: Suppose Cantor's set were countable. Then there would exist a sequence s(n) containing all of them. Consider each element of s(n) in its base 3 representation. As we have established, each digit in each element of the set is either a 0 or a 2, and is between 0 and 1. We can represent each element of s(n) by its own string of ternary digits: s(1): 0.a1a2a3a4... s(2): 0.b1b2b3b4... s(3): 0.c1c2c3c4... s(4): 0.d1d2d3d4... And so on, where each letter-number pair is either 0 or 2. Construct a number q, with the first digit (after the ternary point) being the complement of a1, the second digit the complement of b2, the third digit the complement of c3, and so on. By complement, I mean the other digit (the complement of 0 is 2 and vice versa). Notice that, by our construction, q belongs in the Cantor set but is not any element of s(n). But s(n) was assumed to contain all elements in the Cantor set, giving us a contradiction. So no such sequence s(n) can exist, and the Cantor set is uncountable. QED.

[@wren1728]

Let A denote he set of real numbers in [0, 1] with a unique ternary expansion containing only the digits 0 and 2, union the set of real numbers in [0, 1] with non-unique ternary expansions for which one of their given two expansions contains only the digits 0 and 2. From the video, we have that A is equal to the cantor set. We aim to show that A is uncountable, from which it follows that the cantor set is uncountable. The set of naturals N is countable. By cantor's theorem, P(N) is uncountable. Define f : A -> P(N) by f(x) = { n in N : the nth digit of the ternary expansion of x is 2}. Now let y be in P(N). There is a real number with only 0 and 2 in its ternary expansion such that f(x) = y, and we have x in A. Hence f is surjective from A to P(N). So A is uncountable, and the cantor set is uncountable. Edit to clean things up a little.

[@pifdemestre7066]

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, ...

[@extrams0]

this reminds me of that fake proof how pi equals 4 We start with a quarter of a square (the line (0,1),(1,1),(1,0)) (total length= 2) Then we "collapse" these lines so the corner square falls on the inner circle (the line (0,1),(root2,1),(root2,root2),(1,root2),(1,0) ) (total length still is 2) Repeat this process and we'll end up with a shape that looks like a quarter circle; but still have length 2 (this however isn't a quarter circle, as we have infinite & infinite small horizontal and vertical lines)

[@mailboy95]

Proof by contradiction: Using base 3 Assume the Cantor Set is countable. Then we can write it as a bijection with the natural numbers such that n in the natural numbers goes to x_n in the Cantor set. Let x_n=0.a_n_1 a_n_2..... for every n. Now let y be defined as follows y=0.b_1 b_2..... where b_k is 0 when a_k_k is 2 and visa versa. Thus y differs from each x_n in at least one place, and thus is not in our bijection. Thus there cannot be a bijection with the natural numbers so the Cantor Set is uncountable. :)