Cont: Deeper than primes - Continuation 2

[doronshadmi]

The only way to consider them different is to accept each as an ordered set.

Even if we follow after your ordered sets argument, the diagonal ordered set is included in the ordered set (in the cube case) of |N|² distinct members AND it is not included in the ordered set (in the matrix case) of |N|² distinct members.

Since |N|² = |N| it means that the very notion of complete ordered set of |N| members, is involved with contradiction, or in other words, an ordered set with |N| members can't be both complete AND consistent.

The same conclusion holds also for sets, as explained in http://www.internationalskeptics.com/forums/showpost.php?p=10991364&postcount=919, http://www.internationalskeptics.com/forums/showpost.php?p=10991469&postcount=920 and http://www.internationalskeptics.com/forums/showpost.php?p=10990069&postcount=915.

 

[doronshadmi]

A set has no places, just elements.

The term "place" here is understood in terms of abstract distinct existence of a given element in some set, and not in terms of some physical space.

 

[jsfisher]

The term "place" here is understood in terms of abstract distinct existence of a given element in some set, and not in terms of some physical space.

I never said otherwise.

I'm going to drop ordered set. It isn't quite the right term for what you are doing, anyway; it is related, but not the same thing.

Instead, I will go back to this and approach it directly:

...
Now, by using diagonalization we construct the diagonal set {400,1,2,...} which is definitely not one of the different rows
...

Nothing in your matrix construction disallows this so-called diagonal set {400, 1, 2, ...} from being the same set as any of the matrix rows. Yet, you claimed it "is definitely not one of the different rows."

How can you show it is different from, for example, {1, 126, 231, ...}. Your presentation gives only some of the members, but not all. This set might be (1, 126, 231, 400, 2, ...} and include whatever other members it needs to match your diagonal set. Similarly, your diagonal set could continue with exactly the other members needed to match.

 

[doronshadmi]

I never said otherwise.

I'm going to drop ordered set. It isn't quite the right term for what you are doing, anyway; it is related, but not the same thing.

Instead, I will go back to this and approach it directly:



Nothing in your matrix construction disallows this so-called diagonal set {400, 1, 2, ...} from being the same set as any of the matrix rows. Yet, you claimed it "is definitely not one of the different rows."

How can you show it is different from, for example, {1, 126, 231, ...}. Your presentation gives only some of the members, but not all. This set might be (1, 126, 231, 400, 2, ...} and include whatever other members it needs to match your diagonal set. Similarly, your diagonal set could continue with exactly the other members needed to match.

The answer is already given in http://www.internationalskeptics.com/forums/showpost.php?p=10989002&postcount=909 and it holds for distinct sets, which are distinct from each other by their members (where repetition of members is not allowed in any distinct set) , so all we have to do is to show that the diagonal set is different by one member for each distinct set, in order to conclude that it is different from any set in the matrix.

The same hold for ordered sets, and since they are distinct by their order, the diagonal ordered set is different from any ordered set by its order.

 

[doronshadmi]

How can you show it is different from, for example, {1, 126, 231, ...}. Your presentation gives only some of the members, but not all. This set might be (1, 126, 231, 400, 2, ...} and include whatever other members it needs to match your diagonal set. Similarly, your diagonal set could continue with exactly the other members needed to match.

Since any given set in the matrix is different than the rest of the sets by its members, the same principle holds for the diagonal set, yet it is clearly not one of the sets of the matrix, and things are not going to be changed by adding diagonal sets to the matrix.

The common principle of difference by members matters, and a diagonal set is constructed by this common principle.

In other words, the completeness of the matrix does not hold, where the completeness of the cube holds since it is easy to show that any possible diagonal set, is already some member of another matrix in the cube.

The set of |N|² distinct sets (by their members) is the same set, whether it is arranged as a cube of matrices or as a single matrix.

The rest of the needed details are already given in http://www.internationalskeptics.com/forums/showpost.php?p=10988946&postcount=908 and http://www.internationalskeptics.com/forums/showpost.php?p=10990069&postcount=915 .

 

[zooterkin]

Since any given set in the matrix is different than the rest of the sets by its members,

Since? You haven't shown how you are generating these sets, so how do we know they are different?

 

[doronshadmi]

all we have to do is to show that the diagonal set is different by one member for each distinct set, in order to conclude that it is different from any set in the matrix.

One may claim that if one of the sets in the matrix is N, one can't define a diagonal set that has a member that is different than the rest of the members of set N.

In that case "different by one member" means that the diagonal set is some proper subset of N, such that at least one of the members of N is not a member of the diagonal set.

 

[doronshadmi]

Since? You haven't shown how you are generating these sets, so how do we know they are different?

They are generated by a common principle that guarantees that each set is different from the rest of the sets by its members.

This is the beauty of the power of mathematical abstraction, you do not have to construct objects in details in order to support the validity of a given mathematical argument.

Actually, most sets of infinitely many elements are indistinguishable of each other without using a common principle (at the "level" of some single set, or at the "level" of more than one set).

For example let's take the following infinite sets:

{1, 457438593, 243, 444, ...}

{34, 10, 347, 222, ...}

There is no way to determine if this is one set or two different sets (since order is not considered) without a common principle, which guarantees that each set is different by its members, and this common principle is not established by detailed construction of infinitely many elements.

 

[doronshadmi]

Let's clarify the fact that a given diagonal set of |N|² distinct members across |N|² distinct sets, is different from each one of these |N|² distinct sets.

Here is an example of a set with |N|² distinct sets, where the first set is set N, and the other sets are different proper subsets of it.

       *--------- |N| ---------*
       |                       |

 *--   {1    ,2     ,3    , ...}
 |
       {2    ,400   ,324  , ...}
|N|[SUP]2[/SUP]
       {3    ,565   ,187  , ...}
 |
 *-- ...

Now, by using diagonalization we construct a diagonal set {400,1,2,...}, as follows:

       *--------- |N| ---------*
       |                       |

 *--   {[COLOR="Blue"][B]400[/B][/COLOR]  ,2     ,3    , ...}
 |
       {2    ,[COLOR="Blue"][B]1[/B][/COLOR]     ,324  , ...}
|N|[SUP]2[/SUP]
       {3    ,565   ,[COLOR="Blue"][B]2[/B][/COLOR]    , ...}
 |
 *-- ...

In this example {400,1,2,...} is a proper subset of N, where the blue members are at least one of the members that do not appear in each intersected proper subset.

As about the intersection of the diagonal set with set N (the first set in the example) the intersected member of the diagonal set appears in N, but since the diagonal set is a proper subset of N, it is different than N.

So in the matrix case, the diagonal set is not any one of the |N|² distinct sets.

If the diagonal set is N, it is defiantly different than any of its |N|² distinct proper subsets.

----------------------

If the |N|² distinct sets are arranged as a cube, any diagonal set across a given matrix of that cube is already some set in another matrix of distinct sets of that cube.

In that case one may claim that since the distinct sets of each matrix in the cube are only a part of the |N|² distinct sets, the diagonal set across the single matrix case must be different than any diagonal set across some matrix in the cube, because the blue members (the members of a given diagonal set) are at least one of the members that do not appear in each intersected proper subset of a given matrix, and there are intersections in the single matrix case that do not exist in any diagonal set across some matrix in the cube.

Since the blue members (the members of a given diagonal set) are at least one of the members that do not appear in each intersected proper subset of a given matrix, there is plenty of room to construct exactly the same diagonal set even if only part of the distinct sets appear in some matrix in the cube, and also because |N|² = |N|.

Also the order of the members is not considered in order to determine the distinction among sets (including any diagonal set).

 

[jsfisher]

The answer is already given in http://www.internationalskeptics.com/forums/showpost.php?p=10989002&postcount=909

Hardly. In that post you referred to set members by ordinal position. So, are we full circle now and you are (again) admitting the sets are ordered?

...and it holds for distinct sets, which are distinct from each other by their members (where repetition of members is not allowed in any distinct set)

You really need to stop with these irrelevant parenthetical asides, especially when they are wrong. There is nothing in ZFC Set Theory (as just one example) that disallows repetition of members. In ZFC, {1, 1, 2, 2, 3, 3} and {1, 2, 3} are both sets, and they happen to be identical sets. There is no mechanism in ZFC to distinguish the two.

...so all we have to do is to show that the diagonal set is different by one member for each distinct set, in order to conclude that it is different from any set in the matrix.

Ok, do it. And do it without assigning order to member positions. So, for example, show that {400, 1, 2, ...} from your oft-repeated example is different from {1, 2, 3, ...}.

 

[doronshadmi]

You really need to stop with these irrelevant parenthetical asides, especially when they are wrong. There is nothing in ZFC Set Theory (as just one example) that disallows repetition of members. In ZFC, {1, 1, 2, 2, 3, 3} and {1, 2, 3} are both sets, and they happen to be identical sets. There is no mechanism in ZFC to distinguish the two.

In your example only {1, 2, 3, ...} is a member of the set of |N|² distinct sets, which are distinct of each other only by their members (order is not considered).

(B.T.W, by ZFC {1, 1, 2, 2, 3, 3} and {1, 2, 3} has the same cardinality, which it 3)

Ok, do it. And do it without assigning order to member positions. So, for example, show that {400, 1, 2, ...} from your oft-repeated example is different from {1, 2, 3, ...}.

Already done at the posts after zooterkin's question, which you preferred to not reply to them.

 

[doronshadmi]

Hardly. In that post you referred to set members by ordinal position. So, are we full circle now and you are (again) admitting the sets are ordered?

What is wrong in http://www.internationalskeptics.com/forums/showpost.php?p=10991664&postcount=921 ?

 

[doronshadmi]

EDIT:

I have looked more carefully about the diagonal set across the single matrix case of |N|² distinct sets.

This diagonal set can't be but the set N itself, across the set of all of its proper infinite subsets.

Since according to ZFC, for example, {1,1,1,...,2,2,2,...,3,3,3,...} and {1,2,3,...} are identical sets, only N itself is not a member of the set of all of its proper infinite subsets.

In this case, if we prove that the set of all proper infinite subsets of N has cardinality > |N| we actually prove that there is more than one infinite cardinality.

Without doing it, we are left (in this case) with only one distinct set (set N itself) that is not included in the set of all the proper infinite subsets of N, and in this case |N|+1=|N|.

 

[jsfisher]

EDIT:

I have looked more carefully about the diagonal set across the single matrix case of |N|² distinct sets.

Your matrix has |N| sets. Each row is a set; there are |N| rows.

This diagonal set can't be but the set N itself, across the set of all of its proper infinite subsets.

It doesn't have to be. Your construction only requires that the diagonal set differ from the "n-th" row set in its "n-th" position. (This is an ordering-like relationship based on an unstated mapping function <-- a hidden assumption!! Horrors!!!)

 

[doronshadmi]

It doesn't have to be.

Wrong. The diagonal set must be set N exactly because the set of distinct rows (which is actually the set of all proper infinite subsets of set N (let's call it set P)) includes also all the proper subsets of set N that each one of them does not include exactly one member of set N.

Set P includes also all the proper infinite subsets of set N that each one of them does not include more than one member of set N.

So the diagonal set is at least of the form {1,1,1,...,2,2,2,...,3,3,3,...}, but since according to ZFC {1,1,1,...,2,2,2,...,3,3,3,...} and {1,2,3,...} are identical sets (which means that|{{1,1,1,...,2,2,2,...,3,3,3,...}, {1,2,3,...}}|=1), only N itself is not a member of the set of all of its proper infinite subsets, and the involved equation is |N|+1=|N|, unless one proves that |P|>|N|.

Your construction only requires that the diagonal set differ from the "n-th" row set in its "n-th" position.

No, my construction only requires that the diagonal set differ from each considered row by at least one member that is included in the diagonal set, but it does not included in each considered row. Order is not involved here, simply because set N definitely different than any member of set P (N is not a proper infinite subset of itself).

This is an ordering-like relationship ...

What is exactly ordering-like relationship?

... based on an unstated mapping function <-- a hidden assumption!! Horrors!!!

What do you wish to say here, exactly?

 

[jsfisher]

...delete for now...

 

[jsfisher]

Wrong. The diagonal set must be set N exactly because the set of distinct rows (which is actually the set of all proper infinite subsets of set N (let's call it set P)) includes also all the proper subsets of set N that each one of them does not include exactly one member of set N.

Emphasis added.

Whoa! Stop. You are claiming the set of rows of the matrix includes all of the proper infinite subsets of the set N?

You will need to prove that.

 

[doronshadmi]

Whoa! Stop. You are claiming the set of rows of the matrix includes all of the proper infinite subsets of the set N?

You will need to prove that.

It is certainly can't be done by ZFC since according to ZFC {1,1,1,...,2,2,2,...,3,3,3,...,...} and {1,2,3,...} are identical sets (which means that|{{1,1,1,...,2,2,2,...,3,3,3,...,...}, {1,2,3,...}}|=1).

It meas that ZFC can't be considered as the foundation of Mathematics.

You also have missed

unless one proves that |P|>|N|.

So what axiomatic formal system of sets has to be used in order to prove or disprove it?

Do you have any suggestions?

(b.t.w I have found http://projecteuclid.org/download/pdf_1/euclid.rml/1204834739 interesting)

 

[jsfisher]

Whoa! Stop. You are claiming the set of rows of the matrix includes all of the proper infinite subsets of the set N?

You will need to prove that.

It is certainly can't be done by ZFC since according to ZFC {1,1,1,...,2,2,2,...,3,3,3,...,...} and {1,2,3,...} are identical sets (which means that|{{1,1,1,...,2,2,2,...,3,3,3,...,...}, {1,2,3,...}}|=1).

It cannot be proven under ZFC, but not for the reason you give.

It meas that ZFC can't be considered as the foundation of Mathematics.

This does not follow. Exactly unlike how |{2,2,2}| = 1 follows from the axioms of ZFC.

You also have missed

unless one proves that |P|>|N|.

No, but you seem to have missed that Mathematics works the exact opposite. You do not get to assume something unless and until someone can prove the contrary.

So what axiomatic formal system of sets has to be used in order to prove or disprove it?

Do you have any suggestions?

I've always favored ZFC (although the C part does cause me some discomfort), but you are the one with the assertion. How about you tell us how you intend to prove that the cardinality of the set of all infinite subsets of N has the same cardinality as N.

 

[doronshadmi]

It cannot be proven under ZFC, but not for the reason you give.

So what is the reason ?

This does not follow. Exactly unlike how |{2,2,2}| = 1 follows from the axioms of ZFC.

It follows because ZFC is an example of classical Cantorian set theory (see http://projecteuclid.org/download/pdf_1/euclid.rml/1204834739 page 322) where any axiomatic system with multisets is a non-classical set theory.

According to what is written in this article a pre-set theory has to be developed in order to deal with classical and non-classical set theories under one framework.

In other words, classical-only or non-classical-only set theories can't be considered as the foundation of Mathematics.

No, but you seem to have missed that Mathematics works the exact opposite. You do not get to assume something unless and until someone can prove the contrary.

You seem to ignore the fact that classical-only set theory can't be considered as the foundation of Mathematics.

I've always favored ZFC (although the C part does cause me some discomfort), but you are the one with the assertion. How about you tell us how you intend to prove that the cardinality of the set of all infinite subsets of N has the same cardinality as N.

It is the set of all proper infinite subsets of N, and the diagonal set is actually a multiset, where each proper infinite subset of N is not a mutiset, so a theory that deals with both sets and mutisets has to be developed in order to solve this problem.