Cont: Deeper than primes - Continuation 2

[jsfisher]

• Each function F_i maps S to P(S). (The function has three indexes, the first is the number of the matrix, the second is the number of the column (or row, if the matrices are flipped and then rotated ccw by 90^o) and the third is the number of the place in a given column (or row))

You actually mean two indices. I.e., _iF_j: S -> P(S), where i would be the "matrix index" and identifies a plane of the cube and j would be a column index within a plane. The column is then the codomain of the function.

As a matter of precision, too, the indices are not numbers, since that would restrict them to being countably. Each index is better thought of as an element of S. So, for example, for S = {a,b,c,...}, _aF_b(c) would be the third element in the second column of the first plane. (I'm using the ordinals in an imprecise, colloquial sense.)

Everything is unordered, too. If you want to impose an order, than your argument takes on serious damaged.

• The cardinality of the set of functions is the same as that of S. (yes)
• The functions are injective. (no, they are bijective in terms of infinite composition of functions)

The question was about each function individually. Given the other constraints, they each better be injective. We will revisit your bijective claim later.

• The functions are constructed such at {} is not in the codomain of any function. (the functions are constructed such that exactly some one member of P(S) is not in the codomain of any function)

Again, I was referring to the functions individually. I believe you are referring to the composite of all the functions. And by "some one member" you must mean "exactly one member".

Functions being what they are, without loss of generality we can assume that exactly one member is {}.

• For i <> j, the intersection of the codomains of F_i and F_j is empty (three indexes are needed (as explained in (1)) so this question is not well articulated)

Two indices, as explained earlier. And even at that, the second index is purely for convenience of identifying location in the cube. One index is sufficient to distinguish among the |S| functions.

• Every element of P(S) except {} appears n the codomain of F_i for some i. (Every element of P(S) (except exactly some one particular element of P(S)) appears in the codomain of F where each place in the codomain is indexed by three numbers, as explaines in question number 1)

Ok, so we have |S| functions whose codomains populate the columns in a cube. And there are |S| codomains of |S| members each, so there is a total of at most |S| x |S| = |S| members in the union of all those codomains.

How, exactly, are all the members of P(S) (except for exactly one member) all supposed to appear among the |S| members of the codomain union if |S| < |P(S)|?

|S| x |S| x |S| x ... x |S| = |S|, so go to any higher dimension cube you like. You'll still be short for covering all (except one) members of P(S).

 

[doronshadmi]

As a matter of precision, too, the indices are not numbers, since that would restrict them to being countably. Each index is better thought of as an element of S. So, for example, for S = {a,b,c,...}, _aF_b(c) would be the third element in the second column of the first plane. (I'm using the ordinals in an imprecise, colloquial sense.)

Everything is unordered, too. If you want to impose an order, than your argument takes on serious damaged.

For the matter of precision, since countably and uncountably are based on the assertion that |S|<|P(S)| by using diagolanalizaion, and since any codomain is indexed exactly by three elements of S (where the element in the codomain is some single element of P(S)), it is easy to use order without any damage to my argument, and clearly show that any diagonal set of P(S) elements, is already an element of a given cube, where there is no index for the the cubes because they can't be unioned, since it is impossible to establish the set of all P(S) members that no one of them is mapped with some S member, so the best that we can get is the equation |S|=|S|³+|1|, for all |P(S)| cubes, and being closed under |S| is insufficient in order to conclude that |S|<|P(S)|.

In other words, dianonalaization is insufficient in order to prove that |S|<|P(S)|, and since countably and uncountably are based on the assertion that |S|<|P(S)| by using diagolanalizaion, countably and uncountably itself is no more than an assertion, in the context of what is called Cantor's standard proof to his (what is called) theorem.

 

[jsfisher]

it is easy to use order without any damage to my argument, and clearly show that any diagonal set of P(S) elements, is already an element of a given cube

Great!! Clearly show it.

You'll need to clearly show how each cube element is determined, though. So far, your examples are a bit vague.

You'll need to clearly show how you construct your diagonal set, too. You've been a bit vague on that, too.


It still boils down to your claim that |S| functions are sufficient to have each and every (except exactly one) element of P(S) appearing in the codomain of some function. And that you did by simply assuming |S| = |P(S)| and then using the assumption to "prove" you can't prove |S| < |P(S)|.

 

[doronshadmi]

You'll need to clearly show how each cube element is determined, though. So far, your examples are a bit vague.

Each cube element is determined by its ability to determine the single P(S) element, that is not an element of a given cube, as demonstrated (without a loss of generality) in http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843.

You'll need to clearly show how you construct your diagonal set, too. You've been a bit vague on that, too.

A concrete example (without loss of generality) is given in http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843.

The diagonal set {{a,d},{b,d},{c,d},...} (where {d,d} is not its member) in the example of a second matrix, is already included in the {{}}_cube as some row of |S| |P(S)| in some matrix, that is not a second matrix of {{}}_cube.

As about the indexes of any disjoint cube, it goes as follows (it is bijective in terms of infinite composition of functions):

The indexes of the first matrix:

A --> (A,A,A),(A,B,A),(A,C,A),...
B --> (A,A,B),(A,B,B),(A,C,B),... 
C --> (A,A,C),(A,B,C),(A,C,C),...
...

The indexes of the second matrix:

A --> (B,A,A),(B,B,A),(B,C,A),...
B --> (B,A,B),(B,B,B),(B,C,B),... 
C --> (B,A,C),(B,B,C),(B,C,C),...
...

...

etc. up to |S| matrices for a given cube, where each matrix has its own |S| distinct rows of |S| |P(S)| elements.

And that you did by simply assuming |S| = |P(S)| and then using the assumption to "prove" you can't prove |S| < |P(S)|.

Not at all. I do not assume that |S| = |P(S)|.

I simply show that given any diagonal set with |S| elements, it is already an element of a given cube, out of |P(S)| cubes, where the cubes can't be unioned, since it is impossible to establish the set of all P(S) members that no one of them is mapped with some S member, so the best that we can get is the equation |S|=|S|³+|1|, for all |P(S)| cubes, and being closed under |S| is insufficient in order to conclude that |S|<|P(S)|.

Again, my argument is about the impossibility of diagonalization to prove that |S|<|P(S)|.

In other words, |S|<|P(S)| may be true, but it can't be proven by diagonalizations along matrices, where each matrix belongs to a given cube with |S| matrices, that their common property is the particular |P(S)| member that is not included in any one of them (and this is exactly the meaning of |1| in the equation |S|=|S|³+|1|, for all |P(S)| cubes).

 

[jsfisher]

The indexes of the first matrix:

A --> (A,A,A),(A,B,A),(A,C,A),...
B --> (A,A,B),(A,B,B),(A,C,B),... 
C --> (A,A,C),(A,B,C),(A,C,C),...
...

Exactly how would the "diagaonal set" be derived? I'd be included to use (A,B,C,D),(A,B,C,D),(A,B,C,D),....

 

[doronshadmi]

Exactly how would the "diagaonal set" be derived? I'd be included to use (A,B,C,D),(A,B,C,D),(A,B,C,D),....

If we use the first matrix, the diagonal set of |S| P(S) elements (that is already some row in another matrix of a give cube) is determined along the places that are indexed by (A,A,A),(A,B,B),A,C,C),... (without replacing the contents of these places, but simply be different of each content in each indexed place), as follows:

A --> [COLOR="Blue"](A,A,A)[/COLOR],(A,B,A),(A,C,A),...
B --> (A,A,B),[COLOR="Blue"](A,B,B)[/COLOR],(A,C,B),... 
C --> (A,A,C),(A,B,C),[COLOR="Blue"](A,C,C)[/COLOR],...
...

Please be aware of the difference between a given index (for example (A,B,B)) and a given P(S) element.

I'd be included to use (A,B,C,D),(A,B,C,D),(A,B,C,D),....

jsfisher, we are talking about a cube, and every place in a given cube is indexed by exactly three indexes.

Moreover, what you did is to direct a given function infinitely many times to the same place (determined as a codomain) in a given 4-dim structure.

 

[jsfisher]

If we use the first matrix, the diagonal set of |S| P(S) elements (that is already some row in another matrix of a give cube) is defined along the places that are indexed by (A,A,A),(A,B,B),A,C,C),... , as follows:

A --> [COLOR="Blue"](A,A,A)[/COLOR],(A,B,A),(A,C,A),...
B --> (A,A,B),[COLOR="Blue"](A,B,B)[/COLOR],(A,C,B),... 
C --> (A,A,C),(A,B,C),[COLOR="Blue"](A,C,C)[/COLOR],...
...

You just might want to review what is meant by the diagonal argument. What you are showing in blue doesn't qualify.

 

[doronshadmi]

You just might want to review what is meant by the diagonal argument. What you are showing in blue doesn't qualify.

Please be aware of the difference between a given index (for example (A,B,B)) and a given P(S) element.

 

[jsfisher]

Please be aware of the difference between a given index (for example (A,B,B)) and a given P(S) element.

So, you intended (A,B,C) to correspond to the i, j, and x in _iF_j(x). Clarity in communication can be a positive, you know.

We are back to the basics then:

What are the functions you propose to use to populate your cube? Be specific so we can all agree what element corresponds to a given index.

How do you propose to derive your diagonal set for a given matrix within the cube? Again, be specific so we can all agree what elements are in the set.

 

[doronshadmi]

So, you intended (A,B,C) to correspond to the i, j, and x in _iF_j(x).

No. Each place in the codomain (which is a cube) is determined by three indexes, and each place in the cube contains a single P(S) element, which is mapped with a unique S element in the domain , such that the result of the bijection between the domain and the codomain is a single P(S) element that is not in the cube.

It is shown that given any diagonal set with |S| elements, it is already an element of a given cube, out of |P(S)| cubes, where the cubes can't be unioned, since it is impossible to establish the set of all P(S) members that no one of them is mapped with some S member, so the best that we can get is the equation |S|=|S|³+|1|, for all |P(S)| cubes, and being closed under |S| is insufficient in order to conclude that |S|<|P(S)|.

 

[doronshadmi]

We are back to the basics then:

There is no reason to get back to the basics since your questions are fully answered in http://www.internationalskeptics.com/forums/showpost.php?p=10972748&postcount=864.

 

[jsfisher]

No. Each place in the codomain (which is a cube) is determined by three indexes, and each place in the cube contains a single P(S) element, which is mapped with a unique S element in the domain , such that the result of the bijection between the domain and the codomain is a single P(S) element that is not in the cube.

Are you unable to describe with specificity that mapping from indices to P(S) element?

Are you unable to describe specificity how you derive a diagonal set?

 

[doronshadmi]

Are you unable to describe with specificity that mapping from indices to P(S) element?

Are you unable to describe specificity how you derive a diagonal set?

jsfisher, concrete examples without loss of generality (including all the needed explanations) are given to you "on a silver platter" in http://www.internationalskeptics.com/forums/showpost.php?p=10972748&postcount=864 and http://www.internationalskeptics.com/forums/showpost.php?p=10972917&postcount=866 for both questions.

No one in the universe can externally force you the eat from (and actually understand) what is on this "silver platter" simply because actual understating can't be achieved by externally forcing it on one's mind.

 

[jsfisher]

jsfisher, concrete examples without loss of generality (including all the needed explanations) are given to you "on a silver platter" in http://www.internationalskeptics.com/forums/showpost.php?p=10972748&postcount=864 and http://www.internationalskeptics.com/forums/showpost.php?p=10972917&postcount=866 for both questions.

Nowhere in those posts do you define what each cube index maps to in P(S). That is an important detail to your claim all except one element of P(S) is in the image of the map.

Also, nowhere in those posts do you define how you'd generate a diagonal set. That is also an important detail to you claim.

 

[doronshadmi]

Nowhere in those posts do you define what each cube index maps to in P(S). That is an important detail to your claim all except one element of P(S) is in the image of the map.

Also, nowhere in those posts do you define how you'd generate a diagonal set. That is also an important detail to you claim.

jsfisher, concrete examples without loss of generality (including all the needed explanations) are given to you "on a silver platter" in http://www.internationalskeptics.com/forums/showpost.php?p=10972748&postcount=864 (including its link http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843) and http://www.internationalskeptics.com/forums/showpost.php?p=10972917&postcount=866 for both questions.

EDIT:

In order to be clear in terms of your notation, _aF_b(c) is actually _{[index for a matrix in a cube]}F_{[index for a column in a matrix]}([index for a cell in a column]), where infinite composition of functions are determined from S to P(S), such that |P(S)| cubes can't be unioned, since it is impossible to establish the set of all P(S) members that no one of them is mapped with some S member, so the best that we can get is the equation |S|=|S|³+|1|, for all |P(S)| cubes, and being closed under |S| is insufficient in order to conclude that |S|<|P(S)|.

 

[jsfisher]

jsfisher, concrete examples without loss of generality (including all the needed explanations) are given to you "on a silver platter" in http://www.internationalskeptics.com/forums/showpost.php?p=10972748&postcount=864 (including its link http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843) and http://www.internationalskeptics.com/forums/showpost.php?p=10972917&postcount=866 for both questions.

You were the one that alleged your cube covered all but exactly one element of P(S).

Prove it.

You can't, of course. You have never been able to prove any of your creative assertions, but you should at least try.

 

[doronshadmi]

Prove it.

Already done in http://www.internationalskeptics.com/forums/showpost.php?p=10974779&postcount=875.

 

[jsfisher]

Already done in http://www.internationalskeptics.com/forums/showpost.php?p=10974779&postcount=875.

Assertion is not proof.

 

[doronshadmi]

Assertion is not proof.

No. What is written in http://www.internationalskeptics.com/forums/showpost.php?p=10974779&postcount=875 is a proof.

The intuitive-only step based on diagonalization in terms of matrix is no more than an assertion about |S|<|P(S)|, in case that S is an infinite set.

Actually http://www.internationalskeptics.com/forums/showpost.php?p=10972775&postcount=865 clearly demonstrates that you don't have (yet) the needed understanding in order to conclude any meaningful thing about http://www.internationalskeptics.com/forums/showpost.php?p=10974779&postcount=875.

Moreover, you actually ignored my reply in http://www.internationalskeptics.com/forums/showpost.php?p=10972917&postcount=866 about http://www.internationalskeptics.com/forums/showpost.php?p=10972775&postcount=865, and the same attitude is used by you about http://www.internationalskeptics.com/forums/showpost.php?p=10974779&postcount=875.

Currently diagonalization, as determined by Cantor along a given matrix, is like a religious dogma for you, and as long as this is your basic attitude about the considered subject, there is no use to continue the discussion with you.

 

[doronshadmi]

I think I have found the mistake in my cube argument:

In case of a cube a diagonal matrix is determined across the cube, and the set of its elements is definitely not in the cube, so diagonalization holds in any dimension and can be used in order to conclude that |S|<|P(S)|.

By this reasoning the diagonal set of 4-dim set is a 3-dim set, etc. add infinitum, yet the common principle of all sets is that they are composed mathematical objects.