Cont: Deeper than primes - Continuation 2

[doronshadmi]

Why not? You haven't said how you constructed this matrix,

Simply |N| sets with |N| members each, such the each set is different than the other sets.

or the "diagonal set".

The diagonal set is constructed such that it is different than anyone of the |N| sets by at least one member, and sets do not have members' repetitions.

If we follow the reasoning of transfinite cardinality then |N| rows + one more row = |N| rows, no matter how many rows are added to the |N|*|N| matrix.

So according to this reasoning diagonalization is insufficient in order to determine different transfinite cardinalities.

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As about the cube, it can be constructed as |N|² rows, and in this case repetitions are not allowed, yet any given diagonal set is already some row in that cube, exactly as already shown in the case of columns in http://www.internationalskeptics.com/forums/showpost.php?p=10979865&postcount=890.

So also in this case diagonalization is insufficient in order to determine different transfinite cardinalities.

 

[jsfisher]

Simply |N| sets with |N| members each, such the each set is different than the other sets.

Ok, so the only constraint is that each row's set is different from the any other set in the matrix. Easy enough to meet, but what's the point? You have not identified any purpose for this matrix.

The diagonal set is constructed such that it is different than anyone of the |N| sets by at least one member, and sets do not have members' repetitions.

You'd need to provide an algorithmic method for doing that. But, again I ask, what's the point? Just constructing a set from the diagonal of your matrix doesn't get you any where near the proof method that uses diagonalization.

If we follow the reasoning of transfinite cardinality then |N| rows + one more row = |N| rows, no matter how many rows are added to the |N|*|N| matrix.

I don't think anyone here is surprised that adding a countable number of new members to an infinite set doesn't change its cardinality.

So according to this reasoning diagonalization...

Full stop. You have not incorporated diagonalization into your reasoning, at least not in how in might be used as a proof method, say for example in proving Cantor's Theorem.

 

[doronshadmi]

Full stop. You have not incorporated diagonalization into your reasoning, at least not in how in might be used as a proof method, say for example in proving Cantor's Theorem.

I have incorporated diagonalization into my reasoning as follows:

If diagonalization is used among a given matrix of |N| distinct sets, then the diagonal set is not one of the distinct sets of this matrix.

If diagonalization is used among a given cube of |N|² distinct sets, then the diagonal set is already one of the distinct sets of this cube.

According to transfinite cardinality reasoning |N|² = |N|, and by following this equation, there is a given distinct diagonal set that is included (in the |N|² distinct sets of a given cube) AND it is not induced (in the |N| distinct sets of a given matrix) as one of the distinct members.

Since according to transfinite cardinality reasoning |N|² = |N|, we derived into contradiction, as follows:

A given set (called diagonal set) of |N| natural numbers is a member AND not a member of a set of |N| distinct sets of |N| natural numbers each.

Conclusion: The completeness of the set of all |N| distinct sets of |N| natural numbers each, is not determined, or more generally the completeness of a set with all |N| elements is not determined, which means that Cantor's notion of the completeness of the set of all natural numbers, does not hold (any given natural number is included AND it in not included in what is wrongly called the set of all natural numbers).

As you see, Cantor's very notion of the completeness of the set of all natural numbers does not hold, and since Cantor's theorem (in case of infinite sets) is based on the notion that infinite sets are complete mathematical objects, its theorem does not hold in the case of infinite sets.

 

[jsfisher]

I have incorporated diagonalization into my reasoning as follows:

If diagonalization is used among a given matrix of |N| distinct sets, then the diagonal set is not one of the distinct sets of this matrix.

When and until you provide the actual method for your diagonal set construction, this may become more than just a useless assertion.

Be that as it may, I don't need a matrix and I don't need a diagonal set to construct a set of "|N| distinct" members to which I can identify and add countably many new members without altering the sets cardinality.

You fail to even hint at a result that will be surprising in any way and certainly not damaging to any established result in Mathematics.

...
Since according to transfinite cardinality reasoning |N|² = |N|, we derived into contradiction, as follows:

A given set (called diagonal set) of |N| natural numbers is a member AND not a member of a set of |N| distinct sets of |N| natural numbers each.

Are you alleging your matrix is supposed to contain all possible sets of countably many integers? Your construction of the matrix does not support that assertion.

Without that support, there is no contradiction. With that support, you'd then be showing (assuming you can establish your other assertions as more than just assertions) that your assumption about the matrix led to a contradiction, and so it was the assumption about the matrix that was at fault.

Diagonalization, on the other hand, would continue to stand, unblemished.

 

[doronshadmi]

EDIT:

When and until you provide the actual method for your diagonal set construction, this may become more than just a useless assertion.

I use exactly the same method as used by Cantor.

Be that as it may, I don't need a matrix and I don't need a diagonal set to construct a set of "|N| distinct" members to which I can identify and add countably many new members without altering the sets cardinality.

Your use of countably many (in case of infinite sets) is based on Cantor's diagonal proof of its theorem, which has no formal basis, as clearly seen in http://www.internationalskeptics.com/forums/showpost.php?p=10979865&postcount=890.

You fail to even hint at a result that will be surprising in any way and certainly not damaging to any established result in Mathematics.

What you call "established result in Mathematics" is not based on formal proof in the case of Cantor's standard proof to its (so called) theorem (in the case of infinite sets), but it is based on no more than an intuitive-only step that is not based of any formal\logical reasoning (as clearly seen in http://www.internationalskeptics.com/forums/showpost.php?p=10979865&postcount=890).

Are you alleging your matrix is supposed to contain all possible sets of countably many integers? Your construction of the matrix does not support that assertion.

Without that support, there is no contradiction. With that support, you'd then be showing (assuming you can establish your other assertions as more than just assertions) that your assumption about the matrix led to a contradiction, and so it was the assumption about the matrix that was at fault.

You are still missing the meaning of the equation |N|² = |N|, such that in the case of a matrix with |N| distinct sets, the diagonal set is not included in the matrix, but in the case of a cube of |N|² matrices, the diagonal set is induced as some set of this cube.

Since by transfinite cardinality |N|² = |N| both cases are actually the same set, such that the diagonal set is included AND it is not included in the set of |N| distinct sets of |N| members each.

Your problem is that you simply ignore my cube argument of |N|² matrices and argue only about the single matrix case.

Diagonalization, on the other hand, would continue to stand, unblemished.

Diagonalization actually leads to the contradiction, and as a result the very notion of completeness among infinite sets does not hold, as shown in http://www.internationalskeptics.com/forums/showpost.php?p=10988661&postcount=903.

In other words, your intuition about the completeness of infinite sets is not formally\logically supported and diagonalization is the method that shows exactly why the completeness of infinite sets is not formally\logically supported.

So as you see, I agree with you, Diagonalization would continue to stand unblemished and would continue to show exactly why the completeness of infinite sets is not formally\logically supported.

 

[jsfisher]

I use exactly the same method as used by Cantor.

In that case, you should have no trouble describing the steps one could follow to construct the set.

 

[jsfisher]

Since by transfinite cardinality |N|² = |N| both cases are actually the same set

That does not follow from anything you have presented. You simply asserted it. Without proof, it is of no consequence.


ETA: Added highlighting for focus.

 

[doronshadmi]

That does not follow from anything you have presented. You simply asserted it. Without proof, it is of no consequence.

It is directly follows from the determination that |N|²=|N|, as follows:

The |N|² distinct sets (where each set has |N| members) are arranged as a single matrix, and in this case, the diagonal set is not in the matrix.

The |N|² distinct sets (where each set has |N| members) are arranged as a cube of matrices, and in this case, the diagonal set is already in that cube.

In other words, by using the same set, it is shown that a given set (called diagonal set) is a member AND not a member of the same set, and since |N|²=|N| it means that the very notion of complete set of |N| members, is involved with contradiction, or in other words, a set with |N| members can't be both complete AND consistent.

 

[doronshadmi]

In that case, you should have no trouble describing the steps one could follow to construct the set.

Already done in http://www.internationalskeptics.com/forums/showpost.php?p=10985818&postcount=897, exactly as used by Cantor's construction method (https://en.wikipedia.org/wiki/Cantor's_diagonal_argument) by constructing a set along places _{1,1 2,2 3,3} ... of a given matrix, such that this set has:

A first member that is different than the first member of the first set.

A second member that is different than the second member of the second set.

A third member that is different than the third member of the third set.

etc...

and we get a set of |N| members that is different than each of the sets in the matrix, by one distinguished member.

 

[jsfisher]

A first member that is different than the first member of the first set.

You have already told us, very explicitly, that these are sets. There is no first member.

 

[doronshadmi]

You have already told us, very explicitly, that these are sets. There is no first member.

For any given set of |N| members there can be first, second, third, etc. members.

It does not mean that only one particular member of set with |N| members is placed in the first, second, third, etc. ... place, which means that the order of the places is fixed but not their contents.

So, these are sets.

Moreover, without first, second, third, etc. ... places, diagonalization can't be determined.

 

[jsfisher]

For any given set of |N| members there can be first, second, third, etc. members.

If and only if you impress a total ordering relation. And as soon as the ordering relation is included, they you are no long dealing with sets, but with ordered sets.

It does not mean that only one particular member of set with |N| members is placed in the first, second, third, etc. ... place, which means that the order of the places is fixed but not their contents.

So, these are sets.

So, these are ordered sets.

Moreover, without first, second, third, etc. ... places, diagonalization can't be determined.

Bingo!

 

[doronshadmi]

So, these are ordered sets.

The places are ordered but not the contents.

Still we deal with sets and diagonalization, so what is your point about http://www.internationalskeptics.com/forums/showpost.php?p=10988946&postcount=908?

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Let's simplify what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10988946&postcount=908.

When I say "the same set" I mean that the same diagonal set is included (in case of cube arrangement) AND not included (in case of matrix arrangement), where in both cases the same |N|² distinct sets are involved.

 

[jsfisher]

The places are ordered but not the contents.

So, they are ordered sets. {1, 2, 3, 4, 5,...} and {2, 1, 3, 4, 5, ...} are different sets in your presentation.

 

[doronshadmi]

So, they are ordered sets. {1, 2, 3, 4, 5,...} and {2, 1, 3, 4, 5, ...} are different sets in your presentation.

No, this is the same set in my presentation because the index of a given place and the content of a given place are independent of each other, and so is the case about the diagonal set.

In other words, by using the same set (where each member of this set is distinct from the other members, and for each member the index of a given place and the content of a given place are independent of each other) it is shown that a given set (called diagonal set (where each member of this set is distinct from the other members, and for each member the index of a given place and the content of a given place are independent of each other)) is a member AND not a member of the same set (where each member of this set is distinct from the other members, and for each member the index of a given place and the content of a given place are independent of each other).

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

You may say that in the cube case the diagonal set is determined across a proper subset of the set of |N|² distinct members, where in the matrix case the diagonal set is determined across the whole set of |N|² distinct members.

But it does not matter, since in both cases all we care is that the same diagonal set is included AND not included in the same set of |N|² distinct members, which means (since |N|²=|N|) that a set with |N| members (where each member of this set is distinct from the other members, and for each member the index of a given place and the content of a given place are independent of each other) can't be both complete AND consistent.

 

[jsfisher]

No, this is the same set in my presentation because the index of a given place and the content of a given place are independent of each other, and so is the case about the diagonal set.

You are describing order. As soon as you admitted an index, you have applied order to the set.

A set has no places, just elements.

 

[doronshadmi]

A set has no places, just elements.

Unless indexes are used, yet they are independent of the elements of a given set, and because of this independence, for example, {1, 2, 3, 4, 5,...} and {2, 1, 3, 4, 5, ...} are the same set in my presentation.

The rest of the discussed subject is given in http://www.internationalskeptics.com/forums/showpost.php?p=10990069&postcount=915.

 

[jsfisher]

Unless indexes are used, yet they are independent of the elements of a given set, and because of this independence, for example, {1, 2, 3, 4, 5,...} and {2, 1, 3, 4, 5, ...} are the same set in my presentation.

The indices is what provides the ordering. All you have done is obscured the ordering relation by hiding it behind a mapping function.

Without the ordering, your attempt at diagonalization wouldn't necessarily produce sets different from any set already in the matrix. You require the elements of each row be in some order (an order independent of the ordering of any other row).

Consider this matrix:

{ 1, 2, 3, 4, 5, 6, ... }
{ 1, 3, 4, 5, 6, 7, ... }
{ 1, 4, 5, 6, 7, 8, ... }
{ 1, 5, 6, 7, 8, 9, ... }
...

The set {2, 1, 3, 4, 5, ...} meets your criterion for constructing the diagonal set, yet you claim it is no different from the very first row (which as a set, it of course isn't).

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

 

[doronshadmi]

The set {2, 1, 3, 4, 5, ...} meets your criterion for constructing the diagonal set,

No, it does not meet my criterion for constructing the diagonal set, simply because {2, 1, 3, 4, 5, ...} and {1, 2, 3, 4, 5,...} are the same set since the indexes and the members are independent of each other, so nothing is hidden behind a mapping function.

The reason of the independence is very simple, the indexes of a given matrix are of the form _x,y and the indexes a given cube are of the form _x,y,z .

These forms are not members of any set with infinitely many natural numbers.

 

[doronshadmi]

In addition to my previous post the elements of the indexes are taken from the set K={1',2',3',...} that no one of its members is a member of the set N={1,2,3,...}, so you have no case whatsoever (the Italic style is used in order to clarify that we are talking about the members themselves and not about some notations of them).

In other words, http://www.internationalskeptics.com/forums/showpost.php?p=10990069&postcount=915 survives your ordered sets criticism.