Cont: Deeper than primes - Continuation 2

[jsfisher]

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.

I see. You are absolutely right and I am absolutely wrong. Yep, no use continuing this discussion because you are infallible.

I think I have found the mistake in my cube argument

Oh, wait...you were wrong all along.

+10 points for admitting a mistake. -100 for the arrogance that preceded it.

 

[doronshadmi]

+10 points for admitting a mistake. -100 for the arrogance that preceded it.

Please do not play the teacher on me, I have found the mistake, not you, so please hold your horses.

Anyway I think (from first glance) that I have discovered why a composed domain of distinct elements (if taken in terms of positive whole numbers and 0) is observed as 3d-space, as follows:

1) First we have a non-composed singleton (which is a 0d element), a row of singletons (which is a 1d composed object) and a matrix (which is a 2d composed object).

2) Using a 1d composed diagonal object across a 2d composed object, determines a composed 3d object.

3) Using a 2d composed diagonal object across a 3d composed object, determines a composed 4d object.

4) Using a 3d composed diagonal object across a 4d composed object, determines a composed 5d object.

etc. ... ad infinitum, and we get the triples

0,1,2
1,2,3
2,3,4
3,4,5
...

which give the illusion of 3d-space, where the diagonal composed objects (marked by blue) are the "engine" of these triples.

 

[jsfisher]

Please do not play the teacher on me, I have found the mistake, not you, so please hold your horses.

Your fundamental mistake was and probably continues to be your belief your that an |S| x |S| x |S| cube is big enough to map all the elements of P(S) except for exactly one. It isn't, yet you assert it does in multiple posts, then link back to the assertions even more times, all the while claiming you've proved it.

Let me know when you get past this blunder.

 

[doronshadmi]

Your fundamental mistake was and probably continues to be your belief your that an |S| x |S| x |S| cube is big enough to map all the elements of P(S) except for exactly one. It isn't, yet you assert it does in multiple posts, then link back to the assertions even more times, all the while claiming you've proved it.

Let me know when you get past this blunder.

Now you ignore http://www.internationalskeptics.com/forums/showpost.php?p=10975499&postcount=880 and http://www.internationalskeptics.com/forums/showpost.php?p=10975600&postcount=882.

What a "great teacher" your are.

More generally, new discoveries may be found by challenging the obvious, but not with the help of "great teachers" like you that looks at challenging the obvious as a sign of arrogance.

Moreover, admitting a mistake is the result of actually show the mistake, where challenging the obvious is definitely involved with many mistakes, yet it does not mean that one has to avoid challenging the obvious.

 

[doronshadmi]

Your fundamental mistake was and probably continues to be your belief your that an |S| x |S| x |S| cube is big enough to map all the elements of P(S) except for exactly one.

Wong jsfisher.

1) I did not claim that an |S| x |S| x |S| cube is big enough to map all the elements of P(S) except for exactly one.

2) My mistake is that I used diagonalization in terms of 1d composed objects among a given 3d composed object, in order to claim that diagonalization does not hold among a given 3d composed object.

3) The correction was done by using diagonalization in terms of 2d composed object among a given 3d composed object.

Now I start to examine the notion about diagonalization given in http://www.internationalskeptics.com/forums/showpost.php?p=10975600&postcount=882 as a result of the correction.

 

[jsfisher]

Wong jsfisher.

1) I did not claim that an |S| x |S| x |S| cube is big enough to map all the elements of P(S) except for exactly one.

Hmmm.

... 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 ...

 

[doronshadmi]

Hmmm.

Until this very moment you do not understand my cube argument about diagonalization (which is wrong) which is:

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)|.

My mistake is that I used diagonalization in terms of |S| 1d composed objects among 3d composed object, instead of 2d composed object among 3d composed object.

By using 2d composed object among 3d composed object a given 4d composed object has|S|^|S| P(S) elements and we get the strict inequality |S|<|S|^|S|+1, or in other words, diagonalization holds.

More details are found http://www.internationalskeptics.com/forums/showpost.php?p=10975600&postcount=882.

 

[doronshadmi]

|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).

1) My wrong cube (where a cube is no more than a 3d composed object) arument was about |P(S)| disjoint 3d composed objects, so speaking about higher dimension cube is simply nonsense, or in other words, you were not in any position to understand my mistake about diagonalization in order to correct it, and after I corrected it you do not understand the correction, since you are still thinking in terms of "higher dimension cube" nonsense.

2) An actual higher dimension than a cube is 4d composed object of |S| x |S| x |S| x ... = |S|^|S| P(S) elements, where such 4d composed object can't be determined in terms of |P(S)| 3d composed disjoint cubes (where each disjoint cube has |S|³=|S| elements).

3) As long as your reasoning uses "higher dimension cube" nonsense, you can't understand the content of http://www.internationalskeptics.com/forums/showpost.php?p=10975600&postcount=882.

 

[jsfisher]

Until this very moment you do not understand my cube argument about diagonalization...

You do like to jump around, don't you?

The subject of the post you quoted was contradictory claims you made. It was not about your diagonalization argument nor lack of it.

You told us one of your cubes included all of P(S) except for exactly one member. Do you now retract the claim?

You told us you didn't make the above claim even though your own words show that you did. Do you now retract the claim?

...or is this going to be another exercise in nonsense like when you were claiming a set is the union of its members?

 

[doronshadmi]

You told us one of your cubes included all of P(S) except for exactly one member.

No.

What I actually said is that since (Any possible diagonal set of |S| P(S) elements is already a given column of |S| P(S) elements in some cube) and since (any given cube is constructed such that a given unique element of P(S) is not included in it) and since (it is impossible to establish the set of all P(S) members that no one of them is mapped with some S member (which means that |P(S)| cubes can't be unioned)), then by using diagonalization we are left with 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, we have showed that diagonalization is insufficient in order to conclude that |S|<|P(S)|.

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

In http://www.internationalskeptics.com/forums/showpost.php?p=10975499&postcount=880 I said that I discovered a mistake in my cube argument, because if we use a diagonal matrix across a cube, no one of it diagonal sets is some column in the cube.

Well, I continued to examine the diagonal matrix, and I have discovered that it is possible to show a diagonal set of such matrix that is already a column in the considered cube, and here is a concrete example (without loss of generality):

1) We construct a cube of |S|³ P(S) elements, where infinite composition of functions are determined from S to P(S), in such a way that a given unique element of P(S) is not included in the cube, and in this example the P(S) element not included in the cube is {}.

2) Here is an example of some matrix (let's call it matrix a) of that cube:

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

3) Now we determine some diagonal set across matrix a (where {} is out of its range), without changing the elements of matrix a:

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

4) We discover that the diagonal set across matrix a, is already some column of matrix b:

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

Generally, diagonal sets across matrices do not change even a single element of these matrices, through diagonalizations.

In other words, the diagonal matrix across a given cube does not hold, and we are back to |S|=|S|³+1 for all |P(S)| cubes, which means that diagonalization is insufficient in order to conclude that |S|<|P(S)|.

 

[jsfisher]

What I actually said is that since (Any possible diagonal set of |S| P(S) elements is already a given column of |S| P(S) elements in some cube) and since (any given cube is constructed such that a given unique element of P(S) is not included in it) and since (it is impossible to establish the set of all P(S) members that no one of them is mapped with some S member)

If you say so, but you also said this:

... 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 ...

Do you wish to retract this claim?

 

[doronshadmi]

If you say so, but you also said this:


Do you wish to retract this claim?

They are both the same claim, since you have troubles to get it, please look at this:

1) We construct a cube of |S|³ P(S) elements, where infinite composition of functions are determined from S to P(S), in such a way that a given unique element of P(S) is not included in the cube, ...

The rest of the details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10979865&postcount=890.

 

[jsfisher]

They are both the same claim, since you have troubles to get it, please look at this...

You previously posted that "there is exactly one and only one member of P(S) that is not in the cube".

You later posted that you never made such a claim.


Please decide which, if either, statement you choose to keep. (Hint: Best choice would be to abandon both since they are both false.)

 

[doronshadmi]

You previously posted that "there is exactly one and only one member of P(S) that is not in the cube".

Yes, each cube has |S|=|S|³ P(S) elements, and each cube (out of |P(S)| cubes) is constructed such that "there is exactly one and only one member of P(S) that is not in the cube".

So what exactly is your problem?

 

[jsfisher]

Yes, each cube has |S|=|S|³ P(S) elements, and each cube (out of |P(S)| cubes) is constructed such that "there is exactly one and only one member of P(S) that is not in the cube".

So what exactly is your problem?

This part: "there is exactly one and only one member of P(S) that is not in the cube".

And this part: "out of |P(S)| cubes".

And the part where you denied you said that first part.

 

[doronshadmi]

This part: "there is exactly one and only one member of P(S) that is not in the cube".

And this part: "out of |P(S)| cubes".

And the part where you denied you said that first part.

I did not denied anything.

Each cube (out of |P(S)| cubes) is constructed such that exactly one and only one member of P(S) is not in each cube.

Each cube is named by the P(S) element that is not included in it, and any possible diagonal set is already a column in some 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 |P(S)| cubes can't be unioned, and as a result we are left with the equation |S|=|S|³+|1|, for all |P(S)| cubes.

In other words, we have shown that diagonalization is closed under |S| and therefore it is insufficient in order to conclude that |S|<|P(S)|.

More details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10979865&postcount=890.

 

[doronshadmi]

Let's play with diagonalization among matrix of |N| elements, for example, this is some |N|*|N| matrix with |N| different rows:

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

 *--   {1    ,126   ,231  , ...}
 |
       {2    ,400   ,324  , ...}
|N|
       {3    ,565   ,187  , ...}
 |
 *-- ...

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

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

 *--   {[COLOR="Blue"][B]400[/B][/COLOR]  ,126   ,231  , ...}
 |
       {2    ,[COLOR="Blue"][B]1[/B][/COLOR]     ,324  , ...}
|N|
       {3    ,565   ,[COLOR="Blue"][B]2[/B][/COLOR]    , ...}
 |
 *-- ...

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.

 

[jsfisher]

Let's play with diagonalization among matrix of |N| elements, for example, this is some |N|*|N| matrix with |N| different rows:

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

 *--   {1    ,126   ,231  , ...}
 |
       {2    ,400   ,324  , ...}
|N|
       {3    ,565   ,187  , ...}
 |
 *-- ...

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

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

 *--   {[COLOR="Blue"][B]400[/B][/COLOR]  ,126   ,231  , ...}
 |
       {2    ,[COLOR="Blue"][B]1[/B][/COLOR]     ,324  , ...}
|N|
       {3    ,565   ,[COLOR="Blue"][B]2[/B][/COLOR]    , ...}
 |
 *-- ...

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.

Using curly brackets and calling them sets does not make them sets. They are sequences. Each row of your matrix is a countable sequence of integers. Your demonstration (cleaned up a bit to be presentable) shows that the set of all possible such sequences is uncountable.

This result is not a revelation.

 

[doronshadmi]

Using curly brackets and calling them sets does not make them sets. They are sequences.

In mathematics, a sequence is an ordered collection of objects in which repetitions are allowed. Like a set, it contains members (also called elements, or terms). The number of elements (possibly infinite) is called the length of the sequence. Unlike a set, order matters,...

( https://en.wikipedia.org/wiki/Sequence )

Each row of your matrix is a countable sequence of integers.

No, each row in my matrix is a set of natural numbers, since order does not matter and repetitions are not allowed.

Your demonstration (cleaned up a bit to be presentable) shows that the set of all possible such sequences is uncountable.

This result is not a revelation.

Now use a cube (|N|³ natural numbers (repetitions are allowed, but it does not matter since |N|³=|N|)), and the diagonal set is already some row in that cube.

So you have no case, whatsoever.

 

[zooterkin]

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

Why not? You haven't said how you constructed this matrix, or the "diagonal set".