Cont: Deeper than primes - Continuation 2

[doronshadmi]

That proves nothing at all related to your set of functions.

http://www.internationalskeptics.com/forums/showpost.php?p=10968368&postcount=839 (your matrix method) clearly demonstrates that you do not understand (yet) the cube method in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835 (your misunderstanding is also seen in http://www.internationalskeptics.com/forums/showpost.php?p=10968562&postcount=840).

Once again, without a detailed reply to the content of http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835, your claim above is not supported.

 

[doronshadmi]

One could prove the diagonal set does not appear as a row of the matrix.

In that case your function is from S to some proper subset of S, and it is not relevant to my cube argument of |S|³ P(S) members (where by this argument it is explicitly determined that for each cube (out of |P(S)| cubes), there is exactly one and only one member of P(S) that is not in the cube, and it is impossible to establish the set of all these P(S) members that no one of them is mapped with some S member. ).

Moreover, your argument is only about a given matrix that is the result of S -> S which is irrelevant to a given cube of P(S) members, which is the result of S -> P(S).

 

[doronshadmi]

You need to prove your assertion. Here, I'll repeat it for you since you continue to talk about other things:

You have asserted there exists a set (of the same cardinality as S) of functions that map S to P(S) such that exactly one member of P(S) is not in the codomain of any of the functions.

Prove this set exists.
​

I'll construct the matrices of the {{}}_cube in a different way, which may help to understand the cube argument (without loss of generality).

An example of a first matrix of {{}}_cube (diagonalization is available):

          *--------- |S| ---------*                                            
          |                       |                                                                         
 *-- a -- [{a}  ,{a,b} ,{a,c}, ...] The vertical elements of the matrix,     
 |                                  are the members of |S| proper subsets of P(S)    
     b -- [{b}  ,{b}   ,{b}  , ...] that provide {} as the member of P(S),  
|S|                                 which is not in the range of any of       
     c -- [{c}  ,{c}   ,{c}  , ...] these proper subsets.                              
 |                                                                             
 *-- ...

An example of a second matrix of {{}}_cube (diagonalization is available):

          *--------- |S| ---------*                                            
          |                       |                                                                         
 *-- a -- [{a,d},{a,b} ,{a,c}, ...] The vertical elements of the matrix,     
 |                                  are the members of |S| proper subsets of P(S)    
     b -- [{b}  ,{b,d} ,{b}  , ...] that provide {} as the member of P(S),  
|S|                                 which is not in the range of any of       
     c -- [{c}  ,{c}   ,{c,d}, ...] these proper subsets.                              
 |                                                                             
 *-- ...

... etc. ... and we get a cube of |S| matrices, where no one of the members of that matrices is {}. Therefore this particular cube is called {{}}_cube where other members of set P(S) are easily demonstrated as elements of {{}}_cube.

If you do not believe me, then please provide any member of set P(S) (except {}) and I will easily show that it is already included in {{}}_cube.

Diagonalization is not available among a given cube, which means that given any diagonal set, it is already included as some proper subset of |S| P(S) members in that cube (for example: 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).

In case that a given diagonal set of |S| P(S) members, does not provide {} within {{}}_cube, it is simply an element of a cube that is not {{}}_cube.

Since it is impossible to establish the set of all P(S) members that no one of them is mapped with some S member, then the best that we can get is the equation |S|=|S|³+|{{}}|, for all |P(S)| cubes, and being closed under |S| is insufficient in order to conclude that |S|<|P(S)|.

 

[jsfisher]

I'll construct the matrices of the {{}}_cube in a different way, which may help to understand the cube argument (without loss of generality).

I understand your "cube argument" just fine. It is derived from a set of functions you allege exist. You will need to show their existence before you can make use of them.

You need to complete Step 1 before you can go to Step 2.

You have asserted there exists a set (of the same cardinality as S) of functions that map S to P(S) such that exactly one member of P(S) is not in the codomain of any of the functions.

Prove this set exists.​

 

[doronshadmi]

You will need to show their existence before you can make use of them.

Their existence is shown in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835.

http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843 proves that only one member of set P(S) is not a member of a given cube.

I understand your "cube argument" just fine.

http://www.internationalskeptics.com/forums/showpost.php?p=10968562&postcount=840, http://www.internationalskeptics.com/forums/showpost.php?p=10968580&postcount=841 and http://www.internationalskeptics.com/forums/showpost.php?p=10968723&postcount=842 clearly demonstrate that your claim has no basis.

For the last time jsfisher, please support your claims one by one in details according to what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835 and http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843.

As long as you are avoiding it, your claims about the discussed subject simply have no basis whatsoever.

So it is up to you, "When you have to shoot...Shoot! Don't talk" (or in your case "don't just wave your hands").

 

[jsfisher]

Their existence is shown in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835.

No, all you did was present something you claim to be an example of the set of functions. You failed to show the set has the required property, namely that exactly one member of P(S) is not in the codomain of any function.

For your convenience:

You have asserted there exists a set (of the same cardinality as S) of functions that map S to P(S) such that exactly one member of P(S) is not in the codomain of any of the functions.

Prove this set exists.​

 

[doronshadmi]

No, all you did was present something you claim to be an example of the set of functions. You failed to show the set has the required property, namely that exactly one member of P(S) is not in the codomain of any function.

For your convenience:

You have asserted there exists a set (of the same cardinality as S) of functions that map S to P(S) such that exactly one member of P(S) is not in the codomain of any of the functions.

Prove this set exists.​

jsfisher, your game is over. Now it is clear that you can't support your claims one by one in details according to what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835 and http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843.

This reply closes my discussion with you on this subject, since there is no use to discuss with hands wavers.

 

[jsfisher]

jsfisher, your game is over. Now it is clear that you can't support your claims one by one in details according to what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835 and http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843.

This reply closes my discussion with you on this subject, since there is no use to discuss with hands wavers.

What an interesting non sequitur. What claim of mine are you challenging? The text you quoted is about your claim, not mine.

It is very curious. You made a claim that you cannot support, so you try to divert attention. This is not unusual for you, but it is still curious behavior.

Since you cannot support your claim, your construction stops there. I welcome your refusal to discuss it any further.

 

[doronshadmi]

What an interesting non sequitur. What claim of mine are you challenging? The text you quoted is about your claim, not mine.

It is very curious. You made a claim that you cannot support, so you try to divert attention. This is not unusual for you, but it is still curious behavior.

Since you cannot support your claim, your construction stops there. I welcome your refusal to discuss it any further.

jsfisher, now you are hands waving to yourself, and you are absolutely right, by hands waving to yourself you cannot support any of your claims about http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835 and http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843.

What claim of mine are you challenging?

This claim, for example

I understand your "cube argument" just fine

More details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10968998&postcount=845.

From your side, no details whatsoever, only hands waving to yourself.

When you stop hands waving to yourself, please let me know and we can continue the discussion, if you wish.

 

[jsfisher]

What claim of mine are you challenging?

This claim, for example

I understand your "cube argument" just fine

Geez. Seriously? You want to focus on a throw-away comment just so you don't have to address the prior issue.

Your so-called cube argument is Step 3.
Your so-called matrix argument is Step 2.
Preceding both of those is your set of functions assertion.

Let's deal with Step 1 before moving on. Again, and for your convenience:

You have asserted there exists a set (of the same cardinality as S) of functions that map S to P(S) such that exactly one member of P(S) is not in the codomain of any of the functions.

Prove this set exists.​

 

[Little 10 Toes]

It seems that the current thing that doron does not understand is that Cantor's Theorem is true.

 

[jsfisher]

It seems that the current thing that doron does not understand is that Cantor's Theorem is true.

Yeah, he's fighting that. Amusingly, though, his misguided efforts to show it may not be true (for some reason he's now stopping sort of "isn't true") depend on Cantor's Theorem being true.

 

[doronshadmi]

Your so-called cube argument is Step 3.

No. It is step1 that proves why diagonalization always determines a given set that is already exists as one of the elements that are mapped by a set of |S| functions.

In other words, http://www.internationalskeptics.com/forums/showpost.php?p=10968998&postcount=845 holds, you are still do not understand the cube argument.

 

[doronshadmi]

Yeah, he's fighting that. Amusingly, though, his misguided efforts to show it may not be true (for some reason he's now stopping sort of "isn't true") depend on Cantor's Theorem being true.

jsfisher, a proof of a given theorem is not part of the theorem, yet a statement is considered as a theorem if at least one proof supports it.

I show (by exactly one step, which is exactly the single equation |S|=|S|³+|1|) that diagonalization does not support it.

Generally, the one and only one step of a given equation is simply its own existence.

 

[jsfisher]

No. It is step1 that proves why diagonalization always determines a given set that is already exists as one of the elements that are mapped by a set of |S| functions.

Just to be clear, here, one element (and possibly more) or exactly one element?

 

[doronshadmi]

Just to be clear, here, one element (and possibly more) or exactly one element?

Any provided diagonal set along a given matrix is already in some cube.

The cubes can't be gathered into a one set (which means that any given cube is disjoint from the other cubes exactly because 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)|) as very simply shown in http://www.internationalskeptics.com/forums/showpost.php?p=10968844&postcount=843 and http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835.

 

[jsfisher]

No. It is step1 that proves why diagonalization always determines a given set that is already exists as one of the elements that are mapped by a set of |S| functions.

That is at least three steps. You have a (1) diagonalization of a (2) cube derived from a (3) set of functions.

Since the set of functions is the base for the rest, let's focus on them.

You have a set of functions, F_i. Which if any of the following statements is false (or not necessarily true) for your set:

1. Each function F_i maps S to P(S).
2. The cardinality of the set of functions is the same as that of S.
3. The functions are injective.
4. The functions are constructed such at {} is not in the codomain of any function.
5. For i <> j, the intersection of the codomains of F_i and F_j is empty.
6. Every element of P(S) except {} appears n the codomain of F_i for some i.

 

[doronshadmi]

That is at least three steps. You have a (1) diagonalization of a (2) cube derived from a (3) set of functions.

No. The set of functions, the matrices and the cubes are exactly the same thing, the equation |S|=|S|³+|1|.

The details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10970775&postcount=856.

 

[jsfisher]

No. The set of functions, the matrices and the cubes are exactly the same thing, the equation |S|=|S|³+|1|.

I see. You dismiss the obvious to protect the absurd.

Ok, let's accept, as you say, they are exactly the same thing. My questions about the functions still stand.

 

[doronshadmi]

I see.

No, you don't see (yet).

Ok, let's accept, as you say,

No jsfisher, do not accept what you don't understand, this is the first principle of actual reasoning.

My questions about the functions still stand.

1. 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))
2. The cardinality of the set of functions is the same as that of S. (yes)
3. The functions are injective. (no, they are bijective in terms of infinite composition of functions)
4. 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)
5. 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)
6. 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)

Please pay attention that there is no index for the number of 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)|.

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