End Points of Cantor Set - Countable?

[NortonH]

I say they are not.

[taeto]

The Cantor set can be described as the set of points left over in the closed interval [0,1] after successively removing the middle third open subset of every remaining interval. Then each removed set has the form (m/3^n,(m+1)/3^n). It leaves endpoints m/3^n and (m+1)/3^n. These are rational. Since \(\mathbb{Q}\) is countable, so is the set of endpoints.

[NortonH]

After each stage of removal the number of end points doubles.

After n stages the number of end points is 2^n

As n approaches infinity this becomes the cardinality of the power set of N does it not?

So that must be uncountable.

[taeto]

It is tempting to think so. But then you would be assuming a kind of `continuity property' that is not true for cardinalities: it is not true that \( \lim_{n\to \infty} 2^n = 2^\infty,\) where you can replace \(\infty\) by \(\aleph_0\) if you feel like it. It is not a coincidence that the \(\infty\) symbol is used in such limits and not \(\aleph_0\).

Indeed, you can clearly count all the endpoints as you proceed. An endpoint never gets destroyed once it has been created. At the stage when it gets created, you assign a unique number between \(2^{n-1}+1\) and \(2^n\) to it, so that all endpoints that are created at that stage get their own number, which they keep forever after. Every endpoint gets created at some stage, hence each gets a number, and all get different numbers.

Edited November 8, 2018 by taeto

[studiot]

5 hours ago, taeto said:

It is tempting to think so. But then you would be assuming a kind of `continuity property' that is not true for cardinalities: it is not true that limn→∞2n=2∞, where you can replace ∞ by ℵ0 if you feel like it. It is not a coincidence that the ∞ symbol is used in such limits and not ℵ0 .

Indeed, you can clearly count all the endpoints as you proceed. An endpoint never gets destroyed once it has been created. At the stage when it gets created, you assign a unique number between 2n−1+1 and 2n to it, so that all endpoints that are created at that stage get their own number, which they keep forever after. Every endpoint gets created at some stage, hence each gets a number, and all get different numbers.

Good to mention Aleph₀ here.

You make your point well  +1

[NortonH]

I am not convinced.

Quote

 all endpoints that are created at that stage get their own number, which they keep forever after

I can make exactly the same statement about the power sets of the sets {1}, {1,2}, {1,2,3} ... as they continue towards aleph null.

So it seems that we have two possible hypotheses.

1. It is possible to map the set of end points to the integers

2. Such a mapping is impossible.

I believe it is not possible so I am now obliged to try and demonstrate that such a mapping is not possible. You will need to try and demonstrate the mapping you believe exists.

Edited November 9, 2018 by NortonH

[taeto]

4 minutes ago, NortonH said:

You will need to try and demonstrate the mapping you believe exists.

I thought that I did just that. Or you do not believe that the \(2^{n-1}\) new endpoints that are created at each stage can be numbered with integers from \(2^{n-1}+1\) to \(2^n\)?

8 minutes ago, NortonH said:

I can make exactly the same statement about the power sets of the sets {1}, {1,2}, {1,2,3} ... as they continue towards aleph null.

You have to think more carefully about the "limit" of those power sets. Their limit has to be the same as just their union. The subsets of \(\mathbb{N}\) that appear in the union of these power sets are precisely those subsets that appear in at least one of the sets \(\{1,2,\ldots,n\}\). Such sets are finite subsets of \(\mathbb{N}\). When you take the limit, you will never produce an infinite subset of \(\mathbb{N}\). And the set of finite subsets of \(\mathbb{N}\) is in fact countable.

[NortonH]

I believe your statement but I just point out that the same argument can be made about the power set of each of the sets I listed. Your argument is not rigorous.

I can see the attraction of thinking that at each stage the sets are finite and countable but I do not see how they are any different from the power set cardinality at each stage.

 

[taeto]

6 minutes ago, NortonH said:

Your argument is not rigorous.

I can see the attraction of thinking that at each stage the sets are finite and countable but I do not see how they are any different from the power set cardinality at each stage.

It is not supposed to be rigorous; it is sufficient that it explains what goes on, because your understanding is slightly off.

You like the example of the power sets of {1}, {1,2}, {1,2,3},... We can stick to that, since it is simpler. Each set {1,...,n} has n elements and its power set has 2^n elements.

The limit of the sequence of sets {1},{1,2},{1,2,3}... is equal to \(\mathbb{N}\). Why? Because every element n of \(\mathbb{N}\) belongs to one of these sets, and to all the following sets as well. The latter is important, if n drops out again after a while, then n would not be in the limit of the sequence. In fact even supposing  that n re-enters again, but drops out again infinitely often, then n will not be in the limit. That is how limits of sequences of sets work, do you agree?

The power set of a set {1,2,...,n} in the sequence contains the 2^n subsets of this set. Consider the infinite sequence of these power sets. Which elements belong to some of the sets in this sequence? They have to be subsets of at least on set {1,2,...,n}. In order to stay in the limit of the sequence they have to be subsets as well of all of the sets {1,2,...,n+1},{1,2,...,n+2}... after that. In our case that condition is obviously satisfied, so it is enough that a subset of \(\mathbb{N}\) is a subset of just one set {1,2,...,n} to guarantee that it will be in the limit. Hence this condition becomes necessary and sufficient for a subset to be an element of the limit of the sequence of power sets.

If you still think that there are infinite sets in the limit of the power sets of {1},{1,2},{1,2,3},... please point to any one in particular, then we can consider whether it is in the limit or not.

[NortonH]

I have  no idea what  \(\mathbb{N}\). means.

I am warming to your idea of the mapping of the end points to the 2^n integers as you described above. So I am almost comfortable with the idea of the end points being countable but I will need to think about the difference between the cardinality of the power sets and the end points.

I will reacquaint myself with the original argument used and then return.

[taeto]

2 hours ago, NortonH said:

I have  no idea what  N . means.

Good point. I mean the set of natural numbers (without zero): \(\{1,2,3,\ldots\}.\)

[wtf]

This para in the Wiki article on the Cantor set may shed some light.

Quote

It may appear that only the endpoints of the construction segments are left, but that is not the case either. The number ¹⁄₄, for example, has the unique ternary form 0.020202… = 0.02 [Overbar over .02 didn't reproduce]. It is in the bottom third, and the top ninth of that third, and the bottom twenty-seventh of that ninth, and so on. Since it is never in one of the middle segments, it is never removed. Yet it is also not an endpoint of any middle segment, because it is not a multiple of any power of 1/3

So there are lots of points in the Cantor set that are not endpoints of some middle third. What a tricky set. 

Also, "today I learned." Cantor didn't discover this set! It was discovered by Henry John Stephen Smith.

 https://en.wikipedia.org/wiki/Cantor_set

Edited November 9, 2018 by wtf

[taeto]

The wiki article also has Norton's "argument" for uncountability of the endpoints. Do not trust everything that you read there.

The left over points are those real numbers in [0,1] that have representations in base 3 with 0 and 2 as the only digits. That makes it particularly convenient to prove that the set of left over points is uncountable; changing every digit 2 in the base 3 representation to a 1 defines a bijection with the binary representation of the whole interval [0,1]. E.g. 1/4 represented in ternary by .020202... maps to .010101... which is 1/3 represented in binary.

The endpoints are precisely those points that are represented in base 3 with only finitely many 2's and no 1's.     

Edited November 9, 2018 by taeto

[NortonH]

OK, I understand what big N means but for some reason my browser was displaying it as /"mathbb{}/" or something. It has resolved it now so that is clear.

I am happy now that your function 2^n maps end points to naturals so that is OK.

That just means that my conundrum has moved to the contrast between the power set of N and the number of end points. I am happy with Cantors clever argument that the power set cannot be mapped to N so the problem is the cardinality of the sets. The power set has cardinality 2^alep0. Now I can only assume that this has been extrapolated from finite sets where it is 2^n. So that gives me the impression that as n approaches aleph0 we can say that 2^n approaches 2^aleph0. Yet this seems to be where the problem is.

The Cantor set still baffles me. A countable number of cuts produces an uncountable set. Only the end points experience a cut firsthand and yet somehow an uncountable number of points are left behind separated from all other points.

[taeto]

19 hours ago, NortonH said:

The Cantor set still baffles me. A countable number of cuts produces an uncountable set. Only the end points experience a cut firsthand and yet somehow an uncountable number of points are left behind separated from all other points.

That is a very reasonable comment. It brings us to a more familiar scenario. Suppose we started with the set of all real numbers, and then we took away all the rational numbers from it. Then we have \(\mathbb{R}\setminus \mathbb{Q}\). Removing a countable set \(\mathbb{Q}\) from the uncountable set \(\mathbb{R}\) produces another uncountable set, which has all its pairs of distinct points separated. Are you comfortable with that?

[Carrock]

25 minutes ago, taeto said:

That is a very reasonable comment. It brings us to a more familiar scenario. Suppose we started with the set of all real numbers, and then we took away all the rational numbers from it. Then we have R∖Q . Removing a countable set Q from the uncountable set R produces another uncountable set, which has all its pairs of distinct points separated. Are you comfortable with that?

Surely not. In the real numbers there is a full set of real numbers between each two rational numbers.

[studiot]

1 hour ago, Carrock said:

Surely not. In the real numbers there is a full set of real numbers between each two rational numbers.

How can it be full?

A full set would include rationals (which have been removed).

This is one of pecularities of infinite sets.

Between any two rationals there is a real irrational number

and

Between any two real numbers there is a rational number.

Yet there are more irrational numbers than there are rational ones.

 

https://www.google.co.uk/search?source=hp&ei=ojfnW5ayBcOMsAH_naXQAg&q=between+any+two+real+numbers+there+is+a+rational+onumber&oq=between+any+two+real+numbers+there+is+a+rational+onumber&gs_l=psy-ab.3..0i13k1j0i22i30k1l5.2704.20152.0.20460.56.35.0.21.21.0.232.4480.2j29j2.33.0....0...1c.1.64.psy-ab..2.54.4876...0j0i131k1j0i8i13i30k1.0.H9_9tj3QgJU

[taeto]

1 hour ago, Carrock said:

Surely not. In the real numbers there is a full set of real numbers between each two rational numbers.

There is indeed. But the point was that even though the set of rational numbers has a smaller size than the set of real numbers, it is also true for any two distinct real numbers that there are infinitely many rational numbers between them.

[Carrock]

5 minutes ago, studiot said:

1 hour ago, Carrock said:

Surely not. In the real numbers there is a full set of real numbers between each two rational numbers.

How can it be full?

A full set would include rationals (which have been removed).

aleph-one - aleph-null = aleph-one.

By 'full set' I meant a set of the same size as the set o reals i.e. they can be put in one-to-one correspondence.

Analogously, the infinite sets 1,2,3,4..... and 1,1,2,1/2,3,1/3,4,,1/4..... are the same size

2 minutes ago, taeto said:

There is indeed. But the point was that even though the set of rational numbers has a smaller size than the set of real numbers, it is also true for any two distinct real numbers that there are infinitely many rational numbers between them.

My point was that there is not a rational number between every two irrational numbers in any real number set. i.e. points are not in general separated when rational numbers are removed.

e.g. there is no rational number larger than 2exp(1/2) and smaller than the next highest irrational number.

[taeto]

8 minutes ago, Carrock said:

aleph-one - aleph-null = aleph-one.

By 'full set' I meant a set of the same size as the set o reals i.e. they can be put in one-to-one correspondence.

Analogously, the infinite sets 1,2,3,4..... and 1,1,2,1/2,3,1/3,4,,1/4..... are the same size

Then you would have to agree that between any two distinct reals there is also a `full set' of rationals, i.e. as many rationals as in \(\mathbb{Q}\) altogether? 

[Carrock]

1 minute ago, taeto said:

Then you would have to agree that between any two distinct reals there is also a `full set' of rationals, i.e. as many rationals as in Qaltogether? 

No. That would require that the cardinality of the set of rational numbers be the same as the cardinality of the set of irrational numbers. i.e. 2exp(aleph-null) = aleph-null.

[taeto]

15 minutes ago, Carrock said:

e.g. there is no rational number larger than 2exp(1/2) and smaller than the next highest irrational number.

By speaking of the "next highest" irrational number, you appear to assume a well-ordering of irrational numbers. However, the usual and familiar linear ordering of the reals is not a well-ordering. But I think that in the context of the OP, we are supposed to deal with just the usual ordering.

7 minutes ago, Carrock said:

No. That would require that the cardinality of the set of rational numbers be the same as the cardinality of the set of irrational numbers. i.e. 2exp(aleph-null) = aleph-null.

Would you like to see me write out a proof that for any two real numbers x and y with x < y there are infinitely many rational numbers q such that x < q < y, or would you prefer to work out the easy proof for yourself?

[Carrock]

1 minute ago, taeto said:

By speaking of the "next highest" irrational number, you appear to assume a well-ordering of irrational numbers. However, the usual and familiar linear ordering of the reals is not a well-ordering. But I think that in the context of the OP, we are supposed to deal with just the usual ordering.

That was just a casual example. I'm not familiar with the subtleties of well-ordering so I may have got it wrong.

 

25 minutes ago, Carrock said:

My point was that there is not a rational number between every two irrational numbers in any real number set. i.e. points are not in general separated when rational numbers are removed.

This statement is true no matter how the set is ordered. There is no way you can use aleph-null rationals to separate all the 2exp(aleph-null) points on a line.

[studiot]

2 minutes ago, Carrock said:

This statement is true no matter how the set is ordered. There is no way you can use aleph-null rationals to separate all the 2exp(aleph-null) points on a line.

Did you read the (impressive) list of proofs that I linked to disputing this?

[taeto]

9 minutes ago, Carrock said:

That was just a casual example. I'm not familiar with the subtleties of well-ordering so I may have got it wrong.

A well-ordering is pretty much precisely the example of an ordering < such that if x is any given element, there is one particular element y such that x < y, and there is no z for which x < z < y. The usual ordering < of the real numbers does not have this property, since if x and y are any real numbers with x < y, then z = (x+y)/2 does lie between x and y and is different from both.