Cont: Deeper than primes - Continuation 2

[doronshadmi]

And now it is wrong.

Edit:

Indeed jsfisher, this is the best conclusion that one can get by using intuitive-only basis ,which actually claims that by |S|=|S|*|S|+|1| for all matrices, it is possible to prove that |S|<|P(S)|.

 

[doronshadmi]

Let's correct the argument of {}_matrix, by change it into {}_cube , as follows:

 A matrix of P(S) elements is |S|*|S|, called {}_matrix                 
                                                                               
          *--------- |S| ---------*                                            
          |                       |                                            
                                                                               
 *-- a -- [{a}  ,{b}   ,{c}  , ...]                                            
 |                                                                             
     b -- [{a,b},{b}   ,{c}  , ...]                                            
|S|                                                                            
     c -- [{a,c},{b}   ,{c}  , ...]                                            
 |                                                                             
 *-- ...

By following the example above there is a diagonal ur-element ( for example, [{a,b},{b,c},{c,d},...] ) along this {}_matrix of ur-elements

          *---------- |S| ----------* 
          |                         | 
                                    
 *-- a -- [[COLOR="Blue"][B]{a,b}[/B][/COLOR] ,{b}   ,{c}   , ...] 
 |                                  
     b -- [{a,b} ,[COLOR="Blue"][B]{b,c}[/B][/COLOR] ,{c}   , ...] 
|S|                                 
     c -- [{a,c} ,{b}   ,[COLOR="Blue"][B]{c,d}[/B][/COLOR] , ...] 
 |                                  
 *-- ...

which is not one of any of the horizontal |S| ur-elements of this matrix, yet its elements provide {}, so no matrix is sufficient in order to provide all the ur-elements that provide {} as some member of P(S) that is outside the range of all |S| members of set S, and we have no choice but to extend the {}_matrix into {}_cube (|S|*|S|*|S|, OR |S|³).

In this case the diagonal method is insufficient in order to provide some ur-element that is not one of the ur-elements of a given |S|³ cube.

So, for any #_cube (where # is a placeholder of some P(S) member) |S|=|S|³+|1| holds, and so is the case of about |P(S)| numbers of such cubes.

In other words, what is called Cantor's standard proof for what is called Cantor's theorem, is limited to |S|=|S|³+|1|, and therefore can't prove that |S|<|P(S)|.

 

[doronshadmi]

If I interpret this word salad correctly, Doron, you are trying to say the H = {{}} is the set of elements in P(S) not mapped by any mapping S -> P(S).

Is that right?

It is false, of course, but first we need to understand exactly what you were trying to say.

what I say is as follows:

S is an infinite set.

Any single member of H is some member of P(S), and its identity is determined by a function from all |S| members of S to a given proper subset with |S| members of P(S), where the order of S members w.r.t to the given proper subset with |S| members of P(S), is significant (otherwise the identity of the single H member is not determined).

|S| functions from set S to |S| distinct proper subsets of set P(S) (where each distinct proper subset of set P(S) has exactly |S| members of set P(S)), enable to determine a diagonal function, which provides the distinct single member of H, even if this diagonal function is not one of the |S| functions from set S to |S| distinct proper subsets of set P(S).

So |S| functions from set S to |S| distinct proper subsets of set P(S) (where each distinct proper subset of set P(S) has exactly |S| members of set P(S)), is not enough, and we use |S|² functions from all |S| members of S to |S|² proper subsets with |S| members of P(S) each, where the order of S members w.r.t any proper subset with |S| members of P(S) is significant (otherwise the identity of the single H member is not determined).

|S| proper subsets of set P(S) with |S| members each, is actually a matrix of |S|² P(S) members, and any given matrix provides is sufficient framework for diagonalization.

In that case we are using |S|² proper subsets of set P(S) with |S| members each, which is actually a cube of |S|³ P(S) members.

Since no infinite cube is a insufficient framework for diagonalization (every diagonal function along some matrix of that cube, is already in that cube) what is called the standard proof of (what is called) Cantor's theorem, is closed under the equation |S|=|S|³+|H| (where H has no more than a single member of P(S) for each cube) for all |P(S)| cubes.

|P(S)| disjoint |S|=|S|³+|H| equations (where S is an infinite set) can't be used in order to conclude that |S|<|P(S)|, so Cantor's standard proof is actually based on intuitive-only step that actually can't formally\logically prove that |S|<|P(S)|.

Again, we are talking only about diagonalization.

If Cantor's theorem (in case of infinite sets) is proved without using diagonalization, such that it is not closed under the equation |S|=|S|+1, then this is another story.

 

[doronshadmi]

Let's correct some mistakes in the previous post:

Instate of:

"|S| proper subsets of set P(S) with |S| members each, is actually a matrix of |S|² P(S) members, and any given matrix provides is sufficient framework for diagonalization."

it has to be:

"|S| proper subsets of set P(S) with |S| members each, is actually a matrix of |S|² P(S) members, and any given matrix provides a sufficient framework for diagonalization."


Also instead of:

"Since no infinite cube is a insufficient framework for diagonalization (every diagonal function along some matrix of that cube, is already in that cube) what is called the standard proof of (what is called) Cantor's theorem, is closed under the equation |S|=|S|³+|H| (where H has no more than a single member of P(S) for each cube) for all |P(S)| cubes."

it has to be:

"Since no infinite cube is a sufficient framework for diagonalization (every diagonal function along some matrix of that cube, is already in that cube) what is called the standard proof of (what is called) Cantor's theorem, is closed under the equation |S|=|S|³+|H| (where H has no more than a single member of P(S) for each cube) for all |P(S)| cubes."

 

[doronshadmi]

So let's summarize what it is formally\logically shown so far:

S is an infinite set.

Given |S|² proper subsets of |S| members of P(S) for each proper subset, such that any function from S to each proper subset provides exactly the same one P(S) member that is not in the range of any function (this exactly the same one P(S) member is a member of a set called H), we get cube of |S|³ P(S) members, where the diagonalization does not hold (since any diagonalization along some matrix of that cube, is already in that cube).

A given single and distinct H member (which is some member of P(S)), is provided by |S|² functions from S to |S|² proper subsets of P(S), which are already in the cube of |S|³ P(S) members.

So each cube is notated by exactly by one unique member of P(S), as follows:

{}_cube, {a}_cube, {b}_cube , ... , {a,b,c,...}_cube, where these unique single members of P(S) can't be gathered into the set of all P(S) members that are not mapped with any member of set S, simply because it is guaranteed that, for example (without loss of generality) {} that is not mapped with any S member under {}_cube, is definitely mapped with some S member under any cube that is not {}_cube.

Since diagonalization does not hold among cubes and since it is impossible to establish the set of all P(S) members that are not mapped with any member of set S, we are left with |P(S)| |S|=|S|³+1 disjoint equations, which can't be used in order to conclude that |S|<|P(S)|.

 

[jsfisher]

what I say is as follows:

S is an infinite set.

Any single member of H is some member of P(S), and its identity is determined by a function from all |S| members of S to a given proper subset with |S| members of P(S)

Any function or did you have a particular one in mind. Why must it be a proper subset of P(S)? There is a hidden assumption (and I know how you hate hidden assumptions) that P(S) is larger than S.

...where the order of S members w.r.t to the given proper subset with |S| members of P(S), is significant (otherwise the identity of the single H member is not determined).

So, you need ordered sets for this?

Be that as it may, how does the function determine set H?

 

[doronshadmi]

Any function or did you have a particular one in mind.

S is an infinite set.

I am talking about |S|² functions from S to |S|² proper subsets of P(S) (where each proper subset has |S| members) that provide exactly one member of P(S), which is not a member of any these |S|² proper subsets of P(S).

The |S|² proper subsets of P(S) are arranged as a cube of |S| matrices, where each matrix has |S|² members of P(S), and each diagonal proper subset of P(S) (that has |S| members) that in not mapped with S in a given matrix of that cube, is mapped with S in another matrix (out of |S| matrices) of that cube, or in another cube (in case that the function from S to a given diagonal proper subset of P(S), provides a single member of P(S) that is a member of at least one of the |S|² proper subsets of P(S) of a given cube).

So, given any cube of |S|³ P(S) members, any given diagonal proper subset of |S| P(S) members, is already in some cube, or in other words, we are closed under |P(S)| equations, where each equation is |S|=|S|³+|H| (where for any cube there is exactly one member of H that is not any one of the P(S) members of that cube) or |S|=|S|³+|1|.

|S|=|S|³+|1| can't be used in order to conclude that |S|<|P(S)|.

Why must it be a proper subset of P(S)? There is a hidden assumption (and I know how you hate hidden assumptions) that P(S) is larger than S.

There is no hidden assumption here because:

1) there is a bijection form an infinite set to its proper subset.

and

2) |S|=|S|³+|1| can't be used in order to conclude that |S|<|P(S)|.

and

3) It is impossible to establish the set of all P(S) members that no one of them is mapped with some S member.

So, you need ordered sets for this?

For any cube (where diagonalization does not hold among cubes) in order to get some P(S) member that is not in that cube, the order of S members w.r.t |S|² proper subsets of P(S) members (where each proper subset has |S| members) is not changed in all |S|² functions.

Also the |S| members of each proper subset of P(S) members in a given cube, are ordered in such a way w.r.t S members, which enables to determine the P(S) member that is not in that cube.

Be that as it may, how does the function determine set H?

The answer was given above, and here is some example (without loos of generality):

 *-- a -- [{a}  ,{a,b} ,{a,c}, ...]   The vertical elements of the matrix,     
 |                                    are P(S) members of |S| sets,    
     b -- [{b}  ,{b}   ,{b}  , ...]   which provide {} as the member of P(S)  
|S|                                   that is not a member of any of       
     c -- [{c}  ,{c}   ,{c}  , ...]   these sets (where each set is a    
 |                                    proper subset of set P(S)).                                                                           
 *-- ...                                                                       
                                                                               
                                                                               
                                                                               
 The matrix of these P(S) elemnts is |S|*|S|, called {}_matrix                 
                                                                               
          *--------- |S| ---------*                                            
          |                       |                                            
                                                                               
 *-- a -- [{a}  ,{b}   ,{c}  , ...]                                            
 |                                                                             
     b -- [{a,b},{b}   ,{c}  , ...]                                            
|S|                                                                            
     c -- [{a,c},{b}   ,{c}  , ...]                                            
 |                                                                             
 *-- ...

and there are |S| {}_matrices in a given {}_cube, where {} is a member of P(S) that is not found in the cube, and it is exactly the single member of H in that particular cube.

 

[jsfisher]

S is an infinite set.

I am talking about |S|² functions from S to |S|² proper subsets of P(S) (where each proper subset has |S| members) that provide exactly one member of P(S), which is not a member of any these |S|² proper subsets of P(S).

Is |S|² different from |S|?

What makes you think such a set of functions exists?

 

[doronshadmi]

Is |S|² different from |S|?

No.

What makes you think such a set of functions exists?

This is the whole beauty of transfinite cardinality.

1) there is a bijection form an infinite set to its proper subset (and in this case the set of |S|² functions with its proper subset of |S| functions).

and

2) |S|=|S|³+|1| can't be used in order to conclude that |S|<|P(S)|.

and

3) It is impossible to establish the set of all P(S) members that no one of them is mapped with some S member.

The rest of the details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10967016&postcount=827.

 

[doronshadmi]

Is |S|² different from |S|?

No.

But, for example, the fact that there is a bijection from the set of all natural numbers to its proper subset of even numbers, does not mean that the set of all natural numbers does not exist.

Exactly what holds among the set of all natural numbers and its proper subset of even numbers, holds among the set of |S|² functions and its proper subset of |S| functions.

More generally:

Given an infinite set of functions, its members are not reducible into any one of its infinite proper subsets, even if there is a bijection from this set to any one of its infinite proper subsets.

Because of this irreducibly, cubes of |S|³ hold (where diagonalization does not hold among cubes).

The details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10967016&postcount=827.

 

[jsfisher]

Is |S|² different from |S|?

No.

Then why do you continue to insert it into your posts as if it were important? It isn't.

What makes you think such a set of functions exists?

This is the whole beauty of transfinite cardinality....

I didn't ask about its alleged beauty nor what a bijection would imply nor your obsession with cardinal arithmetic. I asked a simple question, and you did not address it at all.

You have asserted there exists a set 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.

Since you want to use the set of functions in your "proof", you first need to show that such a set of functions exist.

It doesn't, so I am very interested in you proving it does.


ETA: Oh, I forgot to include the restriction on the size of the set of functions: The set of functions has the same cardinality as the set S.

 

[jsfisher]

...delete...

 

[doronshadmi]

ETA: Oh, I forgot to include the restriction on the size of the set of functions: The set of functions has the same cardinality as the set S.

http://www.internationalskeptics.com/forums/showpost.php?p=10967360&postcount=830.

 

[jsfisher]

ETA: Oh, I forgot to include the restriction on the size of the set of functions: The set of functions has the same cardinality as the set S.

http://www.internationalskeptics.com/forums/showpost.php?p=10967360&postcount=830.

So, you skip over the entirety of the post substance to a minor ETA correction and respond with something completely unrelated and irrelevant.

Let's try again:

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]

EDIT:

So, you skip over the entirety of the post substance to a minor ETA correction and respond with something completely unrelated and irrelevant.

http://www.internationalskeptics.com/forums/showpost.php?p=10967360&postcount=830 is not a minor ETA correction, but it is a more detailed reply that is related and relevant to the discussed subject.

Let's try again:

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 you become aggressive, a better way to do that is to say: "Please prove this set exists."

After you become calmer please take your time in order to understand the following very simple example of the construction of such set of |S|² proper subsets of P(S) elements (where each one of them has |S| P(S) members, such that {} is not included in any one of them) that establish a cube of P(S) members, where {} is not one of the elements of that cube (and therefore in this case, the cube is called {}_cube):

An example of the first set of |S| proper subsets of |S| P(S) members each, where the members of the proper subsets are arranged as a matrix (diagonalization is available) where {} is not a member of any proper subset:

          *--------- |S| ---------*                                            
          |                       |                                            
         {                                                                   
 *-- a -- {{a}  ,{b}   ,{c}  , ...},                                            
 |                                                                             
     b -- {{a,b},{b}   ,{c}  , ...},                                            
|S|                                                                            
     c -- {{a,c},{b}   ,{c}  , ...},                                            
 |                                                                             
 *-- ... }

An example of the second set of |S| proper subsets of |S| P(S) members each, where the members of the proper subsets are arranged as a matrix (diagonalization is available) where {} is not a member of any proper subset:

          *--------- |S| ---------*                                            
          |                       |                                            
         {                                                                   
 *-- a -- {{a,b},{b}   ,{c}  , ...},                                            
 |                                                                             
     b -- {{a,b},{b,c} ,{c}  , ...},                                            
|S|                                                                            
     c -- {{a,c},{b}   ,{c,d}, ...},                                            
 |                                                                             
 *-- ... }

... etc. ... and we get a cube of |S| matrices, where no one of the members of the members of that matrices is {}. Therefore this particular cube is called {}_cube (as already written above).

Diagonalization is not available among a given cube, which means that given any diagonalization, it is already included as some proper subset of |S| P(S) members in that cube, or in another cube that is not {}_cube.

So we have demonstrated that given a cube with |S| members (since |S|=|S|³), diagonalization does not hold.

Moreover, 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+1, for all |P(S)| cubes, and |S|=|S|³+1 is insufficient in order to conclude that |S|<|P(S)|.

More details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10967360&postcount=830 (including its link).

 

[jsfisher]

jsfisher you become aggressive, a better way to do that is to say: "Please prove this set exists."

There was nothing at all aggressive about my post. The high-lighting and intention was merely to help focus your attention on the issue before you.

After you become calmer please take your time in order to understand the following very simple example of the construction of such set of |S|² proper subsets of P(S) elements (where each one of them has |S| P(S) members, such that {} is not included in any one of them) that establish a cube of P(S) members, where {} is not one of the elements of that cube (and therefore in this case, the cube is called {}_cube)

You need to show that {} is the only member of P(S) not in any codomain of the functions.

That was your assertion, after all.

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.​

You started the current thread arc with a false assertion about "exactly one". You abandoned that claim, but you substituted another, more convoluted assertion about "exactly one". The current one is false, too. However, I would enjoy greatly your attempt to prove your false assertion. Please give it your best effort.

 

[doronshadmi]

There was nothing at all aggressive about my post. The high-lighting and intention was merely to help focus your attention on the issue before you.

It was aggressive exactly because of omitted the word "Please", simple as that.

You need to show that {} is the only member of P(S) not in any codomain of the functions.

Already done in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835.

You started the current thread arc with a false assertion about "exactly one".

It is proven in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835, for any given cube of |S| P(S) members.

You abandoned that claim,

No, I actually improved it by using cubes of |S| P(S) members.

The current one is false, too.

Please support your claim in details, according to the content of http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835.

Without doing it your "The current one is false, too" is not supported.

 

[jsfisher]

Already done in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835.

That proves nothing at all related to your set of functions. You just continued to assume the set exists and then constructed a matrix from an alleged instance of the set.

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.
​

Since your matrix construction is dependent upon your set of functions actually existing, you need to address the existence proof before we can move on.

 

[jsfisher]

By the way, Doron, while you ponder the proof of that which is false, consider this:

Consider the set of natural numbers, N = {0, 1, 2, 3, ...}.
Consider also the set of functions, F_i(x) = 2ⁱ + x*2ⁱ⁺¹.

Each F_i is an injection from N to N, so the cardinality of the codomain is |N|. It is easy to show that the intersection of the codomains for F_i and F_j for any distinct i and j is empty. It is also easy to show that the only member of N not in any codomain is 0. Lastly, there are |N| such functions.

My set of functions satisfies all of your conditions except that the maps are from N to N rather than N to P(N).

One could construct your matrix from the set of functions and N.
One could generate a diagonal set from that matrix.
One could prove the diagonal set does not appear as a row of the matrix.

So what? What does that prove? And if it proves nothing for an N -> N case, what would it prove were it possible to do for the N -> P(N) case?

 

[doronshadmi]

My set of functions satisfies all of your conditions except that the maps are from N to N rather than N to P(N).

One could construct your matrix from the set of functions and N.
One could generate a diagonal set from that matrix.
One could prove the diagonal set does not appear as a row of the matrix.

Thank you for your example, it demonstrates that you are still missing my cube method, where any given diagonal set already appears in some cube (out of |P(S)| cubes).

Moreover, such cubes can be constructed only by functions form S to P(S) (exactly as done by Cantor), where the method of the construction (as demonstrated in http://www.internationalskeptics.com/forums/showpost.php?p=10967888&postcount=835) determines exactly one and only one member of P(S) that is not included in a given cube.