Talk:Cantor set

message: A Cantor set is a compact perfect set with empty interior. A dynamically defined Cantor set is a Cantor set that may be defined by a family of contacting map (see for example the book by Palis-Takens , page 53).
One may call "The Cantor set" the set defined by  "removing the middle third".
 
Cantor himself defined his set as a perfect set that is nowhere dense. He constructs the thing in several ways. I don't think he even mentions the bit about "removing the middle third" except maybe half way through the paper, as an example, and then, only in passing. At least that's how I remember it going.  
I don't mind that the majority of the article is devoted to a simple example that is pounded to death; but the intro should at least give the correct definition, and note that the "middle third" construction is just an example. linas 05:03, 26 July 2005 (UTC)
replies: text:  Could you give a link to a reference, cause I can't find anything about his original definition online. Well.. if you ever revisit this site.. Fresheneesz 22:04, 3 March 2006 (UTC)
replies: text:  ?? The reference to Cantor's paper is given in "historical references" at the bottom of the article. You can get the book on amazon for about $8 or so, and most university libraries will have it. It is a rather mind-opening read, espcially if all that you know about the Cantor set comes from the crappy descriptions given in pop-lit books on fractals (which is all I knew when I embarked on this journey). I'm guessing most books on topology will also discuss it; I don't think they do the "middle third" contruction either. linas 22:53, 3 March 2006 (UTC)
replies: text: Hmm, maybe I'll look it up at the library here. But I kinda doubt I have enough background on the subject. Why don't you correct this page? Fresheneesz 20:23, 5 March 2006 (UTC)
replies: text: I don't think they do the "middle third" contruction either. Actually, all the books of topology I know do just that. In fact, all sources I know use the term "Cantor set" to denote the "middle third" construction. —Tobias Bergemann 08:31, 6 March 2006 (UTC)
replies: text:  Julian F. Fleron from a Note in Mathematics Magazine, Vol. 67, No 2 (Apr 1994) [stable URL: http://wwww.jstor.org/stable/2690689 ] writes that his first reference to the Cantor ternary set is in a footnote to the statement that “perfect sets need not be everywhere dense” on page 575 of his original paper [referenced below]. She quotes him describing it as: The set of real numbers of the form $$ x= \frac{c-1}{3} + \dots + \frac{c_v}{3^v} + \dots $$ where $c_v$ is $0$ or $2$ for each integer $v$. Zimri2 (talk) 02:40, 17 September 2017 (UTC)

text: Note: Wikipedia has two articles, namely Cantor set and Cantor space. Should they be merged to reduce the confusion? – Tobias Bergemann (talk) 08:15, 24 July 2015 (UTC)
thread_title:  Intro defines Cantor set incorrectly (intro is wrong) 

message: I just thought it was cool to note that in this statement:
The Cantor set can be characterized by these properties: every nonempty totally-disconnected perfect compact metric space is homeomorphic to the Cantor set. 
Despite the massive number of adjectives, (nearly) every one of them had already been linked somewhere above. This is good cross-referencing.
Deco 01:07, 27 Nov 2003 (UTC)
thread_title:  Nice linking 

message: 0                         1/3                       2/3                         1
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Gadykozma 07:02, 15 Jul 2004 (UTC)
thread_title:  Removed textual picture 

message: The image in the page (Cantor_set_in_seven_iterations.png) prints solid black from my (Linux) system, although it displays properly. Has anyone else encountered this? Would there be an objection to the image being modified so that the transparent bits are white? Sasha 17:56, 18 April 2006 (UTC)
thread_title:  Image Problem 

message: Mr. 4 numbers, I didn't understand your addition. Did he have in mind a CONCRETE example of a series diverging on the cantor set (as is implied by the word "particular")? I find it hard to believe, such series were found much later. Maybe you are confusing this with the problem of uniqueness (see set of uniqueness)? What's your reference for this information? Gadykozma 10:55, 29 Jul 2004 (UTC)
replies: text: For now, yes.  Maybe I'll add some things later if I look up specifics that I haven't read about for a long time. Michael Hardy 15:02, 16 Aug 2004 (UTC)

text: How about "discovered" rather than "invented"? Gadykozma 21:36, 24 Sep 2004 (UTC)
thread_title: The trigonometric-series origins of Cantor's set theory

message: Just logging what I have done here.  There were a few problems with the page, which I started fixing and then got carried away.  The overall structure is unchanged, but paras have been rewritten for clarity, completeness and/or correctness.  Summary:
* Prose was "you do this", "if you add up" and the like.  I hope my rewrite is more encyclopædic
* Uncountability: this section went straight into discussion of ternary characterisation of C - I added a preliminary statement as to why you would want to do this.
* Characterisation of set in ternary: made links between ternary construction and existing material on numerals, including illustrative examples.
* Mapping from C to [0,1]: it wasn't explicit that this function is well-defined and onto.  It now is.  I also note that it is not 1-1.
* For nowhere dense, the property proven doesn't quite align with the definition in nowhere dense - fixed.
* Totally disconnected doesn't follow from nowhere dense in general.  (TD is a property of a top space in itself, whereas nowhere dense requires a superset space.)
* added a note that the "middle 80%" version works nicely with decimals
These changes will obviously need some third-party input, or mass-reversion, or whatever.  IMHO now, today, this minute,  I have probably improved it, but I may think differently on re-reading it. Andrew Kepert 08:40, 4 May 2005 (UTC)
thread_title:  Notes on Rewrite, May 2005 

message: the page on null set says the cantor set is a null set. This should be notied if true. Fresheneesz 22:02, 3 March 2006 (UTC)
replies: text: The middle-third construction gives a set with measure zero; any set with measure zero is a null set. However, other constructions, e.g. the Smith-Volterra-Cantor set, is not a null set; you can build Cantor sets with any measure. For example, if you remove the rationals from the reals, you get a set that is is homeomorphic to the Cantor set but has measure one. linas 22:59, 3 March 2006 (UTC)
thread_title: Null set

message: Corrected the reference to Cantor's original paper in Acta Mathematica. Canter 21:23, 16 October 2006 (UTC)
thread_title:  References 

message: In the article:  "The Cantor set is the prototype of a fractal."
Am I right in thinking that most examples (Mandelbrot, etc.) of fractals are continous and that a Cantor set isn't?  This would be a useful consideration to anyone studying fractals.
JWhiteheadcc 11:31, 12 November 2007 (UTC)
thread_title:  Fractals with continous curves? 

message: Using the idea of an arbitary base with one value disallowed for all digits, it is possible to just create an infinite number of sets.
Examples in base-10 (decimal), disallowing 1:
0.023495967029=in the set
0.2121=not in the set
1/9th=0.11111...=not in the set
1/8th=0.125=in the set
Note that 'the set' is to be taken in context as being the set I just defined.
In general, for a base-n set, there are a finite number of sets possible to construct this way.  For decimal, I believe there is 9+9*8+8*7*6+7*6*5*4+6*5*4*3*2 possible solutions.  It might be something else but I'm too tired to double check this.  Please verify.  If instead of a recursion, each cycle uses a different disallowed digit, then there's an infinite number of decimal sets.
I'm wondering if any of this would be useful to add to the encyclopedia.
JWhiteheadcc 11:46, 12 November 2007 (UTC)
replies: text: Looking at the sets generated by removing different digits or different groups of digits is a nice idea for some recreational research. I can think of several interesting questions that would be fun to think about. But I don't think it needs to be in the article, since these various constructions have not yet proven to be of general interest in the areas of mathematics where Cantor sets are commonly studied. — Carl (CBM · talk) 13:18, 12 November 2007 (UTC)
thread_title:  Method to generate arbitary sets 

message: This section mentions the Cantor Set is finite. I'd say it's COUNTABLY finite. Wondering whether to edit this section to add that detail. Might be TMI. :-) Martin Packer (talk) 14:55, 10 December 2007 (UTC)
replies: text: It's not finite and it's not countably infinite; it's uncountably infinite. Michael Hardy (talk) 16:22, 10 December 2007 (UTC)

text: The Cantor set is not countably infinite. It has the same cardinality as the set of real numbers. — Carl (CBM · talk) 20:28, 10 December 2007 (UTC)
thread_title:  What's In The Cantor Set? 

message: It's been a while since I have thought about this, but I am pretty sure if you take out less than 1/3 of each segment, you end up with a set of nonzero measure.  It's a simple geometric series; I don't have time now to go through it but maybe I will add that some other time if no one else is interested.
Uranographer (talk) 01:06, 8 March 2008 (UTC)
thread_title:  Variations of the Cantor Set 

message: The equation for the series sum is given incorrectly as (1/3)*(1-2/3) = 1 ... (1/3)*(1-2/3)=1/9 which is not the sum of the gaps.  The series sum equation should be (1/3)/(1-2/3) = 1.  I have no idea how to show that via math markup, I noodled around but kept getting parsing errors.  Thanks! - goatasaur (talk) 16:04, 25 April 2008 (UTC)
replies: text: Rough fix, markup looks a little weird but at least the equation is right. -- goatasaur (talk) 08:57, 27 April 2008 (UTC)
replies: text: I was wrong, the equation reads 1/3 * 1/(1/3) which is equal to one.  Oops.  -- goatasaur (talk) 15:55, 28 April 2008 (UTC)
thread_title:  "What's in the Cantor Set?" sum equation is wrong 

message: Are 0 and 1 in this set? 118.90.35.242 (talk) 19:42, 12 October 2008 (UTC)
replies: text: Yes. — Carl (CBM · talk) 19:54, 12 October 2008 (UTC)
thread_title:  Layman question 

message: The relation currently given here is true for pretty much any set of points on the real line.  What is needed here is that f_L(C) \cong f_R(C)\cong C =f_L(C) \cup f_R(C) 
Which crucially states that the union of the left and right transformations is precisely the Cantor set.81.134.83.6 (talk) 16:18, 28 February 2018 (UTC)
replies: text: You're raising a good point. You're right that f_L(C) \cup f_R(C)=C but I think that what was actually intended was something like f_L(C) \cong f_R(C)\cong C. I would change it but I'm not sure what's the best way to explain the strong sense in which the sets are homeomorphic. — Carl (CBM · talk) 23:59, 14 October 2008 (UTC)
thread_title:  Self-similarity 

message: obviously the cantor set will contain irrational numbers, but it seems only the end points are left. can some1 gimme an example? cuz i dunno how 2 prove there is one  —Preceding unsigned comment added by Cheat notes (talk • contribs) 12:24, 14 January 2009 (UTC)
replies: text: Nowhere dense means not dense in any open set. Every set is dense in itself. — Carl (CBM · talk) 03:27, 4 March 2009 (UTC)
thread_title:  Irrational numbers 

message: I would love to see a section in this article which discusses applications if there are any.  See the article on eigenvalues and eigenvectors for a really good example of what I'm talking about.  I know this is an important topic without applications, I just think it makes math more accessible to us left-brainers.
4.249.3.9 (talk) 21:30, 2 July 2009 (UTC)
thread_title:  Practical applications 

message: zero set proof is pretty simple but some others would have to be shown as corollaries.
i'm new at adding things to math articles that require computation symbols, how would go about adding them ?Grmike (talk) 03:24, 7 February 2010 (UTC)
thread_title: is it appropriate to add that it's a perfect subset of R, uncountable, it is a zero set ?

message: In the proof that the Cantor Set is uncountable, it says that: 
To see this, we show that there is a function f from the Cantor set C to the closed interval [0,1] that is surjective (i.e. f maps from C onto [0,1]) so that the cardinality of C is no less than that of [0,1]. Since C is a subset of [0,1], its cardinality is also no greater, so the two cardinalities must in fact be equal, by the Cantor–Bernstein–Schroeder theorem.
However, I don't see why it is necessary to bring up the Cantor-Bernstein-Schroeder theorem.  If the Cantor Set's cardinality is greater than or equal to the cardinality of [0,1] and also less than or equal to it, then it follows that they are equal.  If the theorem isn't directly related here, then maybe it should be removed or the reference should be moved somewhere else.  — Preceding unsigned comment added by 68.49.92.23 (talk) 22:35, 10 July 2012 (UTC)
replies: text: The Cantor-Bernstein-Schroeder theorem is the simplest way to take the mathematical meaning of the statements that \gamma \leq \delta and \delta \leq \gamma and to conclude that \gamma = \delta (for cardinals \gamma and  \delta).  Of course, once this theorem is proved, it doesn't need to be proved again.Daqu (talk) 19:21, 24 September 2014 (UTC)
thread_title:  Reference Unnecessary?   

message: Firstly, I'm just a boy interested in maths, so bear with me if i am wrong. But:
I think the first explicit formula of the set is wrong. I think that instead of \lim_{m \to \infty} , there should be a \bigcap_{m=1}^\infty. 
If you agree, I think someone should change it! Otherwise, please correct me.  — Preceding unsigned comment added by 188.183.128.99 (talk) 20:04, 25 May 2013 (UTC)
replies: text: You are not wrong, but \lim_n is acceptable because the sequence of sets is monotone, so it is convergent in some sense: if one associates to each set An in the sequence the function fn equal to 1 on the set An and to 0 outside, then fn(x) converges for every real x to a value f(x) equal to 1 on the "limit set" and 0 outside. Bdmy (talk) 22:50, 25 May 2013 (UTC)
replies: text: Well, I was wrong. The sequence of sets in the article is not monotone, and the formula is really wrong. Bdmy (talk) 23:23, 25 May 2013 (UTC)
thread_title:  Explicit formula 

message: The section Topological and analytical properties contains this statment:
"One can see that Γ is totally disconnected and perfect - thus it is homeomorphic to the Cantor set."
This is very false.  
For example, a countable disjoint union of Cantor sets is also totally disconnected and perfect, but not compact.
Another example is 2(2N). (I.e., the cartesian product of continuum-many copies of the set {0,1} with its discrete topology.) This space is even compact, as well as totally disconnected and perfect — but not homeomorphic to the Cantor set.
Even though I know why the space in question is homeomorphic to the Cantor set, the sentence quoted does not explain it.Daqu (talk) 02:49, 7 September 2014 (UTC)
thread_title:  Massively false statement 

message: The section Construction and formula of the ternary set contains this passage (with the formula omitted):
"Let us note that this description of the Cantor set does not characterize the complement of the Cantor set exactly, since the sets given by the formula [here an explicit formula is given] are not disjoint."
This is nonsense, since the formula given is an explicit formula for the complement of the Cantor set. Thus it characterizes the complement of the Cantor set exactly.
If someone wants to describe the complement as the disjoint union of open intervals, that is fine. But the statement is nonsense.Daqu (talk) 03:07, 7 September 2014 (UTC)
thread_title:  Statement makes no sense 

message: The last paragraph in the section Topological and analytical properties contains this passage:
"For any integer q ≥ 2, the topology on the group G=Zqω (the countable direct sum) is discrete. Although the Pontrjagin dual Γ is also Zqω, the topology of Γ is compact. "
But since the notation Zq is used for more than one thing in mathematics, the meaning of Zq here needs to be specified.
It seems that the intention is not the finite cyclic group Z/qZ, because in that case the countable direct sum of copies of this group is only countable, but its Pontrjagin dual is uncountable.
If on the other hand, the meaning of Zq is the q-adic integers (the only other meaning of Zq that I am familiar with), then this statement seems wrong, also — and in any case, "q-adic integers" makes sense only if q is a prime number — which is clearly is not specified to be.
Also: What does it *mean* where it states: 
"the Pontrjagin dual Γ is also Zqω " 
???  Whatever the intended meaning of Zqω, it is a topological group and for that reason it is unclear what the meaning of "is" is.Daqu (talk) 16:42, 1 March 2016 (UTC)
thread_title:  Confusing statements 

message: Recall that the Cantor set is the set of numbers from interval from 0 to 1 whose base-3 representation does not contain the digit 1. Let's extend it to a set of numbers which contain finitely many digits 1. Start with the Cantor set; note that the set omits the central interval representing numbers which contain the digit 1 at the first position. Let's now allow this interval, as long as the number does not contain any more digits 1; this replaces the omitted interval with a scaled down Cantor set. Now we have three intervals (for numbers with digit 0, 1, or 2 at position 1), each omitting the central subinterval (digit 1 at position 2), which we replace with a scaled down copy of the Cantor set (no more digits 1). Now we have nine intervals (all possible digits at positions 1 and 2), each with an omitted central subinterval which we replace with a scaled sown copy of the Cantor set, etc.
What we get is:
* At layer 0 the Cantor set. (The set of numbers which does not contain the digit 1.)
* At layer 1 one scaled down copy of the Cantor set. (The set of numbers whose last appearance of the digit 1 is at position 1.)
* At layer 2 three copies of the Cantor set. (The set of numbers whose last appearance of the digit 1 is at position 2.)
* At layer 3 nine copies. (Last digit 1 at position 3.)
* At layer 4 ...
So altogether at each layer there are finitely many sets of measure zero, so each layer has itself measure zero. And a union of countably many layers of measure zero still has measure zero. We have now gotten an uncountable, dense subset of interval from 0 to 1 of measure zero.
(Does this set have a name? What are its topological properties?) - Mike Rosoft (talk) 20:56, 24 October 2017 (UTC)
thread_title: Interesting extension of the Cantor set

message: The section with all the properties of the Cantor set would be a lot more readable, if each property were listed before its proof rather than after its proof.  --345Kai (talk) 19:33, 2 May 2018 (UTC)
thread_title:  Suggestions for "Properties" section 

message: I. The sentence in question:
>In arithmetical terms, the Cantor set consists of all real numbers of the unit interval that are expressible as a ternary fraction using only the digits 0 and 2.
You’re trying to say we don’t NEED the digit 1 to express the full set. But the way it's written is poorly conceived and ambiguous. Yes, the numbers CAN all be expressed with 0s and 2s, but they’re not ONLY expressible with 0s and 2s. An infinite amount can ALSO be expressed using the digit 1. Using whimsical notation to express the set with only 0s and 2s is a choice, not a limit of the set, and it’s certainly not an “arithmetical term.”
The following is just as true, ambiguous, and whimsical:
>In arithmetical terms, the set of 3 digit numbers that alternate between 1 and 3 (131, 313) consists of all integers that are expressible as a quaternary sum using only each digit to the power of itself.
As it stands, the sentence only serves to call attention to its own awkwardness and should be rewritten for clarity.
II. From your comment below:
" Its ternary contains only 0s and 2s, but neither the 0 nor the 2 is the repetend. "
This is irrelevant. The point is that the repeating part can only consist of 0s and 2s, but that the non repeating part can also consist of 1s. It has nothing to do with whether the period is equal to or greater than 1.
2604:2000:81CF:2B00:A1AD:998B:217B:9B39 (talk) 13:01, 21 January 2019 (UTC)
thread_title:  repeating fraction with a repetend that uses only 

message: Is it?
::
X1 = {0, 1};
X2 = {0, 00, 01, 1, 10, 11};
X3 = {0, 00, 000, 001, 01, 010, 011, ..., 111};
Xk = ...;
In usual applications the set Xk is the domain of a set of identifiers (or indexes), and k is finite.
There are an intuitive recursive construction rule (illustrated) for each new Xk after X1
: Xk  =  Pk U Xk-1
where Pk is the set of all of 2k numbers expressed as fixed-length (k) bit strings. Example:
:X2  =  P2 U X1  =  {00, 01, 10, 11} U {0, 1}.
The number of elements, |Xk|, after |X1|=2, is the recurrence
: |Xk| = |Xk-1| + 2k.  
Examples of |Xk| for k ranging from 2 to 8:
: |X2| = 2+4=6;   |X3| = 6+8=14;   |X4| = 14+16=30;   ...;   |X8| = 254+256=510.
PS: by induction we see that |Xk| = 2k+1-2.
Krauss (talk) 11:04, 17 March 2019 (UTC)
thread_title:  Please explain the mapping to binary sequences 

message: The Cantor Set  is not only a mathematical abstraction to demonstrate "the infinite", but also a classical model and perhaps the simplest fractal with a complete and consensual documentation. As real-world Cantor Set (a linear Cantor dust produced by light) see e.g.  ref1 or ref2.
   
Because of finite size of the real-world fractal structures, the correct model is the finite Cantor set.  
And perhaps the most didactic and most used abstraction, it is not the Cantor set!  The ref1 named it "hierarchical tree of the Cantor set". The tree Tk use the constructor Ck in a different way... Instead of intersection the hierarchy is an union: T_k = \bigcup_{m=1}^k C_m. We can show also that |T_k| = |T_{k-1}| + 2^k = 2^{k+1}-2.
All concepts, the finite Cantor set and the  hierarchical tree of the finite Cantor set are encyclopedic. It is important to introduce it in the article.  Krauss (talk) 16:55, 24 March 2019 (UTC)
thread_title:  Please add hierarchical tree of the Cantor set  

message: The next-to-last paragraph in the section Topological and analytical properties reads as follows:
"For any integer q ≥ 2, the topology on the group G=Zqω (the countable direct sum) is discrete. Although the Pontrjagin dual Γ is also Zqω, the topology of Γ is compact. One can see that Γ is totally disconnected and perfect - thus it is homeomorphic to the Cantor set. It is easiest to write out the homeomorphism explicitly in the case q=2. (See Rudin 1962 p 40.)"
Can we assume that Zq denotes the group of integers modulo q ???
If so, why are we concerned with the "countable direct sum" of this group with itself???
That is not the Cantor set (the direct sum is only countable), so: What is the connection?
Or perhaps Zq instead denotes the q-adic integers???
Clarification is needed here.47.44.96.195 (talk) 15:14, 28 October 2020 (UTC)
thread_title:  Unclear paragraph 

message: An editor has identified a potential problem with the redirect :Cantor`s dust and has thus listed it for discussion. This discussion will occur at Wikipedia:Redirects for discussion/Log/2022 October 15#Cantor`s dust until a consensus is reached, and readers of this page are welcome to contribute to the discussion.  1234qwer1234qwer4 15:28, 15 October 2022 (UTC)
thread_title:  "Cantor`s dust" listed at [[Wikipedia:Redirects for discussion|Redirects for discussion]] 

message: This is my first section on a talk page, so I do not exactly know what goes in here, and I am judging from other sections. Also, I have only done 10 article edits. You are free to edit this section to more reflect the ideals followed on the talk page.
According to the current version of the article, a Smith–Volterra–Cantor set where the middle 8/10 is removed at each iteration can be represented as all real numbers from 0 to 1 that can be written in base 10 with only 0s and 9s. I think it is possible to generalize this fact to that a Smith–Volterra–Cantor set where the middle (b-2)/b is removed can be represented as all real numbers from 0 to 1 that can be written in base b with only 0s and symbols representing b-1. A corollary would be that the Cantor Set can be defined as all real numbers from 0 to 1 that can be written in base 3 with only 0s and 2s. Here is a non-rigorous proof of this:
Lets suppose a Smith–Volterra–Cantor set where the middle (b-2)/b is removed and all numbers are written in base b.
Cutting off the first middle b-2/b in the first iteration leaves us with intervals [0, 1/b] and [(b-1)/b, 1]
This can be written in base b as [0, 0.1] and [0.(b-1), 1]
Considering the left portion of one of the intervals , we can repeat this process in the second iteration, with that interval of length 1/n, or in base n, 0.1.
The intervals created will be 1/n of the size of the previous ones.
Repeating this indefinitely results in a summation across each interval, and since each corresponds to a digit of the number in base b (either 0 or b-1) because each interval is 1/n of the previous one in the previous iteration, every number in this Smith–Volterra–Cantor set can be represented in base b with only 0s and (b-1)s. AbrarBinCiraj (talk) 13:10, 9 February 2024 (UTC)
thread_title:  Generalization of a mentioned fact