Continuous set and continuum hypothesis

[pengkuan]

This article explains why the cardinality of a set must be either Aleph0 or |ℝ|.

In "Cardinality of the set of binary-expressed real numbers" I have shown that binary numbers cannot fill the unit real interval in spite of infinity of digits. In general, the members of a discrete set can only occupy isolated points in a continuous space leaving empty intervals behind. So, the fundamental difference between a continuum and a discrete set is the continuity, not the number of elements. Continuity and discreteness are collectively exhaustive and mutually exclusive. The cardinality of discrete set is aleph0, that of cintinuous set is |ℝ|. Because a set must be continuous or discrete, its cardinality must be either aleph0 or |ℝ|, but never in between. So, the continuum hypothesis is true.

1. Rational numbers are discrete

2. Real numbers are continuous

3. Collectively exhaustive and mutually exclusive events

4. Continuum hypothesis

5. Cardinality of discontinuous subsets of real numbers

Here is the article in pdf Continuous set and continuum hypothesis.pdf

Please read the article at

PDF Continuous set and continuum hypothesis

http://pengkuanonmaths.blogspot.com/2015/12/continuous-set-and-continuum-hypothesis.html

or

Word https://www.academia.edu/19589645/Continuous_set_and_continuum_hypothesis

 

[uncool]

This article explains why the cardinality of a set must be either Aleph0 or |ℝ|.

 

In "Cardinality of the set of binary-expressed real numbers" I have shown that binary numbers cannot fill the unit real interval in spite of infinity of digits.

No, you haven't. You've asserted this, and have not yet dealt with objections to it. As such, you should focus on addressing that before trying to move on.

In general, the members of a discrete set can only occupy isolated points in a continuous space leaving empty intervals behind. So, the fundamental difference between a continuum and a discrete set is the continuity, not the number of elements.

There is a theorem, however, that a continuum necessarily has uncountably many elements (and, if I remember correctly, that it must have at least as many elements as the real numbers). You are correct that the fundamental difference lies in topology, not in set size.

Continuity and discreteness are collectively exhaustive and mutually exclusive.

That depends on how you define discreteness. By the definition that is used in modern topology (a set is topologically discrete if every point forms an open set by itself), no, they are not exhaustive. If you are going to make this assertion, you will have to define discreteness.

The cardinality of discrete set is aleph0

Depending on your definition, you will either have to prove this, or prove that the conditions are exhaustive and exclusive.

, that of cintinuous set is |ℝ|. Because a set must be continuous or discrete, its cardinality must be either aleph0 or |ℝ|, but never in between. So, the continuum hypothesis is true.

This seems to contradict the above. You said that the fundamental difference isn't the number of elements, but are now saying that the number of elements determines whether a set is discrete or not.

1. Rational numbers are discrete

Not according to the topological definition. They are totally disconnected, but not discrete.

2. Real numbers are continuous

3. Collectively exhaustive and mutually exclusive events

4. Continuum hypothesis

5. Cardinality of discontinuous subsets of real numbers

 

Here is the article in pdf Continuous set and continuum hypothesis.pdf

 

Please read the article at

PDF Continuous set and continuum hypothesis

http://pengkuanonmaths.blogspot.com/2015/12/continuous-set-and-continuum-hypothesis.html

or

Word https://www.academia.edu/19589645/Continuous_set_and_continuum_hypothesis

You don't seem to have studied point-set topology, or basic set theory and logic. If you really want to make these claims, you should study those.

 

You are making assertions based on claims made in other threads that have not been established. As your claims in other threads are actually wrong, your conclusions here are also wrong. Further, your new claims in this thread are also, for the most part, wrong.

Edited December 10, 2015 by uncool

[Strange]

 

This article explains why the cardinality of a set must be either Aleph0 or |ℝ|.

In "Cardinality of the set of binary-expressed real numbers" I have shown that binary numbers cannot fill the unit real interval in spite of infinity of digits.

 

I think what you meant to say is that your supposed proof has been thoroughly debunked. The flaws in almost every step have been carefully explained.

[pengkuan]

 

I think what you meant to say is that your supposed proof has been thoroughly debunked. The flaws in almost every step have been carefully explained.

I'm correcting the errors in it.

[Xerxes]

As a late-comer to the party, here's my bottle....

 

The OP (and indeed the title) sems to confuse the noun "continuum" with the adjective "continuous". They are not related - continuity refers to functions whereas a continuum is taken as an ordered domain that is complete, very loosely indeed this means no "gaps".

 

Try this. Suppose the real numbers [math]\mathbb{R}[/math] ordered in the usual sense. Define subsets [math]U,\,L[/math] such that

1. every [math]u \in U[/math] is an upper bound for the set [math]L[/math].

2. every [math]l \in L[/math] is a lower bound for [math]U[/math]

 

Since [math]U[/math] inherits the order from [math]\mathbb{R}[/math], there must be a least element (say) [math]x \in U[/math]. So [math]x[/math] is a lower bound for [math]U[/math], and therefore, by construction must also be in [math]L[/math]

 

That is, the "cut" that defines [math]U,\,L[/math] passes "through" [math]x[/math]. And since the "cut point" is entirely arbitrary, one may say there are no "gaps" in [math]\mathbb{R}[/math] so the Real numbers are a complete ordered domain (field in fact) and can be referred to as a continuum

Edited December 10, 2015 by Xerxes

[pengkuan]

That depends on how you define discreteness. By the definition that is used in modern topology (a set is topologically discrete if every point forms an open set by itself), no, they are not exhaustive. If you are going to make this assertion, you will have to define discreteness.

 

I have defined discreteness in section 1, surely not in formal language of mathematics.

 

Depending on your definition, you will either have to prove this, or prove that the conditions are exhaustive and exclusive.

 

I cannot prove.

 

This seems to contradict the above. You said that the fundamental difference isn't the number of elements, but are now saying that the number of elements determines whether a set is discrete or not.

 

The fundamental difference is continuity. I use this to prove that discrete set and continuous set have different cardinalities and there is not value in between. Size of the set is a result, not the condition.

 

You don't seem to have studied point-set topology, or basic set theory and logic. If you really want to make these claims, you should study those.

You are right. I haven't studied these matters. I learn in writing and discussing.

 

You are making assertions based on claims made in other threads that have not been established. As your claims in other threads are actually wrong, your conclusions here are also wrong. Further, your new claims in this thread are also, for the most part, wrong.

 

Maybe. But I'm confident that there is something true in my claims.

As a late-comer to the party, here's my bottle....

 

The OP (and indeed the title) sems to confuse the noun "continuum" with the adjective "continuous". They are not related - continuity refers to functions whereas a continuum is taken as an ordered domain that is complete, very loosely indeed this means no "gaps".

 

Try this. Suppose the real numbers [math]\mathbb{R}[/math] ordered in the usual sense. Define subsets [math]U,\,L[/math] such that

1. every [math]u \in U[/math] is an upper bound for the set [math]L[/math].

2. every [math]l \in L[/math] is a lower bound for [math]U[/math]

 

Since [math]U[/math] inherits the order from [math]\mathbb{R}[/math], there must be a least element (say) [math]x \in U[/math]. So [math]x[/math] is a lower bound for [math]U[/math], and therefore, by construction must also be in [math]L[/math]

 

That is, the "cut" that defines [math]U,\,L[/math] passes "through" [math]x[/math]. And since the "cut point" is entirely arbitrary, one may say there are no "gaps" in [math]\mathbb{R}[/math] so the Real numbers are a complete ordered domain (field in fact) and can be referred to as a continuum

 

You have given a brilliant definition of the continuum of the real line that I'm not able to give. Thanks. And thanks for joining. You are not late.

 

Intuitively the real line is a continuous line. So I have borrowed "continuity" to qualify the continuum.

[uncool]

I have defined discreteness in section 1, surely not in formal language of mathematics.

And what is your definition?

 

I cannot prove.

Then you shouldn't assert your statement.

 

The fundamental difference is continuity. I use this to prove that discrete set and continuous set have different cardinalities and there is not value in between. Size of the set is a result, not the condition.

First note: you haven't proven it at all.

 

Second note: If you were correct, then the size would be a fundamental difference between the two.

 

You are right. I haven't studied these matters. I learn in writing and discussing.

That's my point. You learn in writing and discussing, not by blindly asserting. What you are doing is blindly asserting instead of asking questions.

 

Maybe. But I'm confident that there is something true in my claims.

Then learn to prove them.

[Strange]

I'm correcting the errors in it.

 

When you do that, you will find that your claims are false. Which will be an excellent indication that you have learned the maths that you need.

[Xerxes]

Let me try to stop this getting ugly.

 

Pengkuan is confused but at least he is honest - as uncool rightly says he is using terms which properly belong in point set topology. Hell, even the Real line [math]R^1[/math] is a topological construction, but pengkan admits he knows nothing of these matters.

 

That's fine - it is not a crime. Some of us may consider abusing termiology as a crime, but hey, I think now I see what he is saying.

 

Let's use pengkuan's definitions of a "continuous set" versus a "discrete set". So the assertion is that they have different cardinalities.

 

Fine.

 

Let us take my last post as a (semi)proof that the Real numbers [math]\mathbb{R}[/math] are complete. Let's further assume that this is what pengkuan means by "continuous".

 

Here is another denifintion of a complete ordered field [math]\mathbb{F}[/math]. If for every monotonically decreasing sequence in [math]\mathbb{F}[/math] there is a limit in [math]\mathbb{F}[/math] then this field is complete. I assert that [math] \mathbb{R}[/math] is complete is this sense as well as in my last sense i.e. has no "gaps".

 

I call on Cantor to tell me that the cardinality of [math]\mathbb{R}[/math] is uncountable.

 

I now consider the Rationals [math]\mathbb{Q}[/math] as a subset of [math]\mathbb{R}[/math], and consider the sequence of partial sums [math]S_n=\sum \nolimits_{n=0}^{\infty}\frac{1}{n!}[/math] which converges on [math]e[/math] which is not rational.

 

Therefore [math]\mathbb{Q}[/math] is not complete. Let's assume that is what pengkuan means by "discrete"

 

And since any element in the set of rational numbers can be written as the quotient of two natural numbers (excluding 0), and since the natural numbers are essentially by definition countable, then [math]\mathbb{Q}[/math] must also be a countable set.

 

So yes, using pengkuan's clumsy (and wrong) notation, continuous set is uncountable whereas a discrete set is countable. This applies to the Real numbers and the Rational numbers as a subset - whether it applies more genarally, I cannot be bothered to think about just now

 

 

 

 

 

 

[studiot]

Xerxes, go well with your intervention.

 

We have been through a long, exhausting and exasperating thread with the OP on just this subject.

 

http://www.scienceforums.net/topic/92509-cardinality-of-the-set-of-binary-expressed-real-numbers/

[uncool]

There are "incomplete" sets that are uncountable. Take, for example, the set of the irrational numbers. Those are "incomplete" in exactly the same way, but are clearly uncountable.

[pengkuan]

There are "incomplete" sets that are uncountable. Take, for example, the set of the irrational numbers. Those are "incomplete" in exactly the same way, but are clearly uncountable.

Your are right. Please see the section 5. Cardinality of discontinuous subsets of real numbers of my pdf in the original post.

Xerxes, go well with your intervention.

 

We have been through a long, exhausting and exasperating thread with the OP on just this subject.

 

http://www.scienceforums.net/topic/92509-cardinality-of-the-set-of-binary-expressed-real-numbers/

Thanks studiot

Let me try to stop this getting ugly.

 

Pengkuan is confused but at least he is honest - as uncool rightly says he is using terms which properly belong in point set topology. Hell, even the Real line [math]R^1[/math] is a topological construction, but pengkan admits he knows nothing of these matters.

 

That's fine - it is not a crime. Some of us may consider abusing termiology as a crime, but hey, I think now I see what he is saying.

 

Let's use pengkuan's definitions of a "continuous set" versus a "discrete set". So the assertion is that they have different cardinalities.

 

Fine.

 

Let us take my last post as a (semi)proof that the Real numbers [math]\mathbb{R}[/math] are complete. Let's further assume that this is what pengkuan means by "continuous".

 

Here is another denifintion of a complete ordered field [math]\mathbb{F}[/math]. If for every monotonically decreasing sequence in [math]\mathbb{F}[/math] there is a limit in [math]\mathbb{F}[/math] then this field is complete. I assert that [math] \mathbb{R}[/math] is complete is this sense as well as in my last sense i.e. has no "gaps".

 

I call on Cantor to tell me that the cardinality of [math]\mathbb{R}[/math] is uncountable.

 

I now consider the Rationals [math]\mathbb{Q}[/math] as a subset of [math]\mathbb{R}[/math], and consider the sequence of partial sums [math]S_n=\sum \nolimits_{n=0}^{\infty}\frac{1}{n!}[/math] which converges on [math]e[/math] which is not rational.

 

Therefore [math]\mathbb{Q}[/math] is not complete. Let's assume that is what pengkuan means by "discrete"

 

And since any element in the set of rational numbers can be written as the quotient of two natural numbers (excluding 0), and since the natural numbers are essentially by definition countable, then [math]\mathbb{Q}[/math] must also be a countable set.

 

So yes, using pengkuan's clumsy (and wrong) notation, continuous set is uncountable whereas a discrete set is countable. This applies to the Real numbers and the Rational numbers as a subset - whether it applies more genarally, I cannot be bothered to think about just now

 

Xerxes

 

Thanks for intervening.

 

My objective is to prove the continuum hypothesis. So, I distinguish discrete sets and continuum. By my definition of discrete set, all points are isolated. So they are disjoint. By my definition of continuum, all points are in contact with one another. Since points can only be disjoint or in contact, the points are in a collectively exhaustive and mutually exclusive game. So, discreteness and continuity are also collectively exhaustive and mutually exclusive.

 

Because of the above property, sets are only discrete set or continuum. So, there can only be cardinalities of discrete set or of continuum, that is, À₀ or

|ℝ|. So, there cannot be cardinality strictly between À₀ and |ℝ|. Hence, the continuum hypothesis is true.

 

For my definition of discreteness and continuity. Also for discontinuous subset of real numbers. Please see here:

http://pengkuanonmaths.blogspot.fr/2015/12/continuous-set-and-continuum-hypothesis.html

[uncool]

I will not be going to your site. If you want to make an argument, make it here.

 

Do you accept that the irrational numbers are "discrete"?

[pengkuan]

I will not be going to your site. If you want to make an argument, make it here.

 

Do you accept that the irrational numbers are "discrete"?

Here is my article

Continuous set and continuum hypothesis.pdf

[studiot]

 

pengkuan

By my definition of discrete set, all points are isolated. So they are disjoint. By my definition of continuum, all points are in contact with one another

 

Have you come across the idea of interior, boundary and exterior points?

 

Or the idea of a neighbourhood?

Edited December 11, 2015 by studiot

[pengkuan]

 

Have you come across the idea of interior, boundary and exterior points?

 

Or the idea of a neighbourhood?

No, I have not these idea. Do these properties determine continuity or size of sets?

[imatfaal]

!

Moderator Note

Please post abstract/summary of argument here. We do not allow OP to force members off-site in order to be able to follow argument. This applies to request to download and open a PDF file. If you wish to have the discussion here then the basis for argument must be made here.

 

I would also be happier if you could work through the clear errors in your other thread first. There is nothing approaching an agreement that the basis for your argument is correct - let alone that your hypothesis might have merit

[pengkuan]

!

Moderator Note

Please post abstract/summary of argument here. We do not allow OP to force members off-site in order to be able to follow argument. This applies to request to download and open a PDF file. If you wish to have the discussion here then the basis for argument must be made here.

 

I would also be happier if you could work through the clear errors in your other thread first. There is nothing approaching an agreement that the basis for your argument is correct - let alone that your hypothesis might have merit

Sorry. I post below the detail of my point

 

1. Discreteness of set

 

Discreteness is the property that qualifies sets that are formed by isolated points. Rational numbers are a discrete set whose elements are ratios of two natural numbers. For example q₀ and q_{i ,} q_i+1 is the mid point between the former. When i®¥, the number q_i+1 approaches q₀ without reaching it. Figure 1 shows the insertion of q_i+1 between q₀ and q_i. So, however small the interval [q₀, q_i] is, there can be infinity of numbers q_j for j > i such that q₀ < q_j < q_i. As there are numbers between any two rational numbers, neighboring rational numbers are disjoint. This property defines the discreteness of a set. As q₀ and q_i can be any rational number in the real line, the set of rational numbers is discrete over the entire real line.

2. Continuity of set

 

Contrary to the rational numbers, the set of real numbers is continuous, that is, any point on the real line is in contact with its adjacent points. “In contact” means there is no gap between these two points in which another number can be inserted. Figure 2shows 2 adjacent points x and y which are in contact and an interval [x, x+e], e being the width of the interval. Because the gap is zero, however small the width of the interval is, it can never be smaller than y-x. This property defines the continuity of a set. As x can be any point in the real line, the set of real numbers is entirely continuous.

 

3. Collectively exhaustive and mutually exclusive events

 

When tossing a coin, all possible outcomes are heads or tails. The values heads or tails are said to be collectively exhaustive, that is, there is no other possibility. Also, when heads occurs, tails can't occur and vice versa. These two values are said to be mutually exclusive, that is, the outcome is either heads or tails, no mixed value is allowed, for example half heads and half tails.

 

In a well ordered set, two neighboring points are either disjoint or in contact, there is no other possibility. So, disjoint or in contact are collectively exhaustive and mutually exclusive. Disjoint elements make discrete sets and elements that are in contact with one another make continuous set. Thus, discreteness and continuity are collectively exhaustive and mutually exclusive too and sets can only be discrete or continuous, no other possibility is allowed.

 

4. Continuum hypothesis

 

The elements of any infinite discrete set can be put in an indexed list where each element uniquely corresponds to its index. This one-to-one correspondence between the elements and the natural numbers proves that the cardinality of any infinite discrete set is À₀, for example the sets of the natural numbers and the rational numbers have cardinality À₀.

 

The cardinality of a continuous set is |ℝ|, like the real numbers. Because a set must be discrete or continuous, its cardinality must be À₀ or |ℝ| and not strictly between À₀ and |ℝ|. In consequence, the continuum hypothesis is true.

 

5. Cardinality of discontinuous subsets of real numbers

 

Can discontinuous subset of real numbers have cardinality smaller than |ℝ|?

 

The first possibility is to remove all algebraic numbers from the real line. The set of the real numbers depleted of all algebraic numbers is not continuous because the places of the algebraic numbers are holes. The cardinality of the algebraic numbers is À₀. So, the cardinality of this subset is |ℝ|-À₀. As we have |ℝ|-À₀=|ℝ|, the cardinality of this subset is not smaller than |ℝ|.

 

Second possibility is to remove real numbers proportionally, for example, removing 9 points in 10 from the unit interval [0, 1]. This way, it seems that the left points are only 1/10 of the original points. But removing 9 points in 10 is equivalent to removing 9 parts in 10 from the unit interval and the leftover numbers are equinumerous to 1/10 of the unit interval, that is, the interval [0, 0.1]. As the cardinality of [0, 0.1] is also |ℝ|, the cardinality of this subset is not smaller than |ℝ|.

 

Another possibility is Cantor ternary set which is constructed in splitting an interval in 3 and removing the central one, then applying this process to the remaining intervals forever. This process seems to reduce the number of points indefinitely. But it turns out that the cardinality of this set is |ℝ|.

 

Then we can remove all the irrational numbers from the unit interval [0, 1]. This way we are left with the rational numbers whose cardinality is À₀. So, the cardinalities of the above subsets of the real numbers are either À₀ or |ℝ| but not strictly in between.

 

In consequence, construction of discontinuous subsets from the real line cannot give cardinality bigger than À₀ and smaller than |ℝ| and the continuum hypothesis holds.

[studiot]

 

pengkuan

Discreteness is the property that qualifies sets that are formed by isolated points. Rational numbers are a discrete set whose elements are ratios of two natural numbers. For example q₀ and q_{i ,} q_i+1 is the mid point between the former. When i®¥, the number q_i+1 approaches q₀ without reaching it. Figure 1 shows the insertion of q_i+1 between q₀ and q_i. So, however small the interval [q₀, q_i] is, there can be infinity of numbers q_j for j > i such that q₀ < q_j < q_i. As there are numbers between any two rational numbers, neighboring rational numbers are disjoint. This property defines the discreteness of a set. As q₀ and q_i can be any rational number in the real line, the set of rational numbers is discrete over the entire real line.

 

 

 

 

What a pity you did not wait for my thoughts in reply to your post16.

 

I have been preparing one of my 'rough guides' this afternoon.

But this takes a measure of time and effort.

Normally a student of this subject would take around 6 months to cover the material, you say you have not studied, and that I am trying to pull together for you in an afternoon.

Many very clever people from many countries spent 150 years or so pulling and pushing, arranging and rearranging this theory into a shape where it is not only the foundation for a great deal of traditional mathematics but provides a springboard for generalisation beyond the numbers.

 

As such the modern theory is in pretty good shape and does not need a rather flaky competitor.

 

 

 

Talking of measure, you are struggling because you are lacking the idea of measure or distance or metric.

Without this your statement

 

 

pengkuan

Contrary to the rational numbers, the set of real numbers is continuous, that is, any point on the real line is in contact with its adjacent points. “In contact” means there is no gap between these two points in which another number can be inserted.

 

has no meaning.

 

What is a gap?

 

What is touching?

 

What is adjacent?

 

For your information modern theory recognises three types of set points, not two. The three types are as I listed in post 15.

 

Since you have now introduced continuity and connectedness

the terms Complete, Compact, Continuous, Connected and Covering all have special (very carefully defined) meanings in relation to this subject.

 

Just to be going on with until I complete my rough guide and to show that connected (=no gap) is a complicated concept here is an extract courtesy Buck Advanced Calculus.

 

Edited December 11, 2015 by studiot

[uncool]

Sorry. I post below the detail of my point

 

1. Discreteness of set

 

Discreteness is the property that qualifies sets that are formed by isolated points. Rational numbers are a discrete set whose elements are ratios of two natural numbers. For example q₀ and q_{i ,} q_i+1 is the mid point between the former. When i®¥, the number q_i+1 approaches q₀ without reaching it. Figure 1 shows the insertion of q_i+1 between q₀ and q_i. So, however small the interval [q₀, q_i] is, there can be infinity of numbers q_j for j > i such that q₀ < q_j < q_i. As there are numbers between any two rational numbers, neighboring rational numbers are disjoint. This property defines the discreteness of a set. As q₀ and q_i can be any rational number in the real line, the set of rational numbers is discrete over the entire real line.

Capture.JPG

The problem here is that the real numbers also have this property. Between any two distinct real numbers, there is another real number (for example, as with the rational numbers, the average of two real numbers is a real number).

 

So if you want to claim that the real numbers are not discrete, then you must have a different definition of "discrete".

[pengkuan]

The problem here is that the real numbers also have this property. Between any two distinct real numbers, there is another real number (for example, as with the rational numbers, the average of two real numbers is a real number).

 

This is very sharp point. Thank you. I have corrected my definition of discreteness as below.

 

Discreteness is the property that qualifies sets that are formed by isolated points. Isolation means these points are surrounded by points of other nature, like island is isolated by water. In the real line each rational number is surrounded by irrational numbers that isolate them from its rational neighbors. Let p and q be two rational numbers. The gap between them is g=q-p. We can create the irrational number p+g/pi that isolates p from q.

 

When q®p, q never reaches p and the irrational quantity g/pi is never zero. Figure 1 shows the isolation of p from its neighbors q and q_-. As p and q can be any rational numbers in the real line, the set of rational numbers is discrete over the entire real line.

[uncool]

Then you should also accept that the irrational numbers are "discrete". Between any two distinct irrational numbers, there is a rational number.

Edited December 12, 2015 by uncool

[pengkuan]

Then you should also accept that the irrational numbers are "discrete". Between any two distinct irrational numbers, there is a rational number.

Yes. But real numbers are continuous, because between any two reals are only reals.

Then you should also accept that the irrational numbers are "discrete". Between any two distinct irrational numbers, there is a rational number.

The question is, how irrationals are uncountable? Isn't it?

[uncool]

Yes. But real numbers are continuous, because between any two reals are only reals.

 

The question is, how irrationals are uncountable? Isn't it?

Before I talk about the number of irrationals, do you accept that by your definition, the irrational numbers are "discrete"?

[pengkuan]

Before I talk about the number of irrationals, do you accept that by your definition, the irrational numbers are "discrete"?

Yes. I agree. I try to figure out another definition.