Cont: Deeper than primes - Continuation 2

[doronshadmi]

The two trees are identical, except one has a star at the top, the other has a zero.

The right tree does not have a path that starts with bit 1, so it is not identical to the left tree, yet both trees have uncountable paths, where the difference of identities and the non-difference of cardinality, is exactly the essence of my argument.

In order to fully understand my argument, one has to read whole of the following links

http://www.internationalskeptics.com/forums/showpost.php?p=11477539&postcount=2096

http://www.internationalskeptics.com/forums/showpost.php?p=11480114&postcount=2097

http://www.internationalskeptics.com/forums/showpost.php?p=11480438&postcount=2098

http://www.internationalskeptics.com/forums/showpost.php?p=11481363&postcount=2100

before one replies to any part of my argument.

 

[jsfisher]

The two trees are identical, except one has a star at the top, the other has a zero.

Of course, but why should facts interrupt are really convoluted presentation? Besides, the whole tree and path confusion is so Doronshadmi can sneak in a list of uncountable infinitely many elements. This cannot be done, of course, but it is buried in so much other wrong it is easily missed.

 

[doronshadmi]

This cannot be done, of course

These kinds of replies are useless.

Please follow after http://www.internationalskeptics.com/forums/showpost.php?p=11481365&postcount=2101 and my request there, if you really wish to air your view about the issue at hand.

And if you replay, then please do it according to what is explained in whole of these links (which means that you have to read whole of them, before you reply in details and right to the point to any part of them).

I am not going to reply again to any kind of useless replies as done in http://www.internationalskeptics.com/forums/showpost.php?p=11481534&postcount=2102.

 

[jsfisher]

This cannot be done, of course

These kinds of replies are useless.

The post was addressed to zooterkin, and I believe he understood what was written.

Had he not, though, I am very confident he would not have simply said, "These kinds of replies are useless." He would instead have asked for clarification about any part he didn't understand.

While you didn't respond as zooterkin would have had he been in your position, I will infer from the part of my post that you quoted, it was the "cannot be done" part of the post confused you.

In your post, you presented a set of uncountably many elements as if it could be listed. You showed the members as a list. So far so good?

This cannot be done, of course. Uncountable infinite sets cannot be listed. If they could, well, then they wouldn't be uncountable, now would they?

 

[doronshadmi]

Uncountable infinite sets cannot be listed.

EDIT:

The arrangement of an infinite set, whether it is arranged as a tree or as a list, has no influence on its cardinality, which can be (by using the standard terminology) countably infinite or uncountable.

First, I prove that the fact that a list has complements that are not included in it, does not prevent the list to be uncountable.

Second, I prove that an infinite set of natural numbers has more than one infinite cardinality, so the Cantorian notion of one and only one infinite cardinality (called aleph₀) has no logical basis, exactly because I use an infinite binary tree as the logical basis of the issue at hand.

Since aleph₀ has no logical basis, so is the case about 2^aleph₀.

In other words, if one defines numbers by also not using radix point along the paths of the infinite binary tree, then those numbers (except number 000...) are infinite spectrum of infinite cardinalites, where some example of such infinite cardinalites is 1000... > 01000... > 001000... > ...

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

The infinite binary tree is used here without loss of generality, which means that any n>2 valued infinite logical tree can be used in my argument, but instead of complements that are not included in a given infinite set, there are simply distinct paths that are not included in a given set.

In other words, my argument logically holds by using any n>1 valued infinite logical tree, and it holds exactly because I define numbers by directly use logical trees, where logical trees are, by definition, can't be illogical.

 

[jsfisher]

The arrangement of an infinite set, whether it is arranged as a tree or as a list, has no influence on its cardinality, which can be (by using the standard terminology) countably infinite or uncountable.

I await your proof that the members of an uncountably infinite set can be put into a linear list.

First, I prove that the fact that a list has complements that are not included in it, does not prevent the list to be uncountable.

No, first you assumed the elements of an uncountable set could be put into a linear list. They cannot.

As for being able to show there is something not in a specific set is less than trivial. The set of real numbers in the interval [0,1) is an uncountable set, yet it does not include the number 1. This revelation does not impress.

 

[abaddon]

I like your txt illustrations and will be using them in some three months hence.

 

[doronshadmi]

I await your proof that the members of an uncountably infinite set can be put into a linear list.

You are still missing this simple logical fact:

The arrangement of an infinite set, whether it is arranged as a tree or as a list, has no influence on its cardinality, which can be (by using the standard terminology) countably infinite or uncountable.

Take, for example, these two infinite logical trees:

               *                                   0
              / \                                 / \
             /   \                               /   \
            /     \                             /     \
           /       \                           /       \
          /         \                         /         \
         /           \                       /           \
        /             \                     /             \
       /               \                   /               \
       0               1                   0               1
      / \             / \                 / \             / \
     /   \           /   \               /   \           /   \
    /     \         /     \             /     \         /     \
   /       \       /       \           /       \       /       \
   0       1       0       1           0       1       0       1
  / \     / \     / \     / \         / \     / \     / \     / \
 /   \   /   \   /   \   /   \       /   \   /   \   /   \   /   \
 0   1   0   1   0   1   0   1       0   1   0   1   0   1   0   1
/ \ / \ / \ / \ / \ / \ / \ / \     / \ / \ / \ / \ / \ / \ / \ / \
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1     0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
             . . .                               . . .

The left tree is an ordered set of distinct logical connectives from contradiction (000...) to tautology (111...), and there is no problem to define a bijection from N to the left tree.

The right tree is an ordered set of distinct logical connectives from contradiction (000...) to 0111.., and there is no problem to define a bijection from N to the right tree.

Cantor "prooved" that there is no bijection form N to R, by using the fact that there are logical trees that do not include infinitely many logical connectives (Cantor's diagonal argument).

Cantor's diagonal argument does not prevent the logical fact that there is bijection from N to the left logical tree or to the right logical tree.

Moreover, similarly to what is shown here, the fact that |S| < |P(S)| by Cantor's theorem ( https://en.wikipedia.org/wiki/Cantor's_theorem ) does not prevent the fact that S and P(S) are already uncountable, exactly as N and the arbitrary mixed set of distinct paths are already uncountable. So in both cases we can ignore the fact that we can define an element that is not paired with some N (or some S) element.

This revelation does not impress.

I totally agree with you, but Cantor is the person that used such facts in order to show that N can't be (by using the standard terminology) uncountable.

I proved that his very notion of infinite sets was wrong (more details are given in http://www.internationalskeptics.com/forums/showpost.php?p=11482631&postcount=2105) and my prove is directly based on ordered logical connectives that, by definition, can't be illogical.

 

[jsfisher]

You are still missing this simple logical fact

It is not I who is missing a simple fact.

However, you are welcome to prove the impossible: Show how the elements of an uncountable set can be enumerated. Via a tree or a list, it doesn't matter. Have at it.

 

[jsfisher]

...
The left tree is an ordered set of distinct logical connectives from contradiction (000...) to tautology (111...), and there is no problem to define a bijection from N to the left tree.

The right tree is an ordered set of distinct logical connectives from contradiction (000...) to 0111.., and there is no problem to define a bijection from N to the right tree.

You need to be consistent. Either include the root in your path designation, or not, but be consistent.

After that, be so kind as to present your "no problem to define a bijection" piece.

 

[doronshadmi]

You need to be consistent. Either include the root in your path designation, or not, but be consistent

The root is logically not any one of the bits (or more generally, nodes) in a given infinite logical tree that it is ordered from contradiction (logical connective 000...) to tautology (logical connective 111...).

Moreover, without this root, no infinite ordered logical tree from contradiction (logical connective 000...) to tautology (logical connective 111...) can be defined.

Since the right tree is exactly half of the left tree, it seems that its root is bit 0, but also in this case, the question is "half of what?" and answer is "half of the left tree, which its root in not any of the bits".

So in both cases we are dealing with the same reasoning.

 

[doronshadmi]

It is not I who is missing a simple fact.

However, you are welcome to prove the impossible: Show how the elements of an uncountable set can be enumerated. Via a tree or a list, it doesn't matter. Have at it.

Already done in http://www.internationalskeptics.com/forums/showpost.php?p=11483171&postcount=2108.

Every distinct ordered logical connective in the left or right infinite ordered logical tree is paired with exactly one N member, such that no member of both mapped sets (one mapping is from the left tree to N, and the other mapping is from the right tree to N) is left out.

One can't argue with logic itself as used in the issue at hand.

 

[doronshadmi]

Show how the elements of an uncountable set can be enumerated.

I show exactly the opposite which is:

N is (by using the standard terminology) uncountable exactly as the left or right infinite ordered logical trees are (by using the standard terminology) uncountable.

 

[jsfisher]

Every distinct ordered logical connective in the left or right infinite ordered logical tree is paired with exactly one N member, such that no member of both mapped sets (one mapping is from the left tree to N, and the other mapping is from the right tree to N) is left out.

This would be the part you need to prove rather than simply assert.

 

[doronshadmi]

This would be the part you need to prove rather than simply assert.

Both sets are perfectly ordered, so there is no problem to define a bijection from each distinct N member to each distinct member of the left or right (by using the standard terminology) uncountable infinite ordered logical trees.

This is the straightforward logic itself that does not need any further mathematical work on the issue at hand.

Moreover, similarly to what is shown here, the fact that |S| < |P(S)| by Cantor's theorem ( https://en.wikipedia.org/wiki/Cantor's_theorem ) does not prevent the fact that S and P(S) are already (by using the standard terminology) uncountable, exactly as N and the arbitrary mixed set of distinct paths are already (by using the standard terminology) uncountable. So in both cases we can ignore the fact that we can define an element that is not paired with some N (or some S) element.

Cantor's diagonal argument has no logical basis, exactly because he have missed the fact that missing members from infinite ordered logical trees, is not a proof that there is no bijection from these (by using the standard terminology) uncountable sets to N.

I await your proof that the members of an uncountably infinite set can be put into a linear list.

However, you are welcome to prove the impossible: Show how the elements of an uncountable set can be enumerated. Via a tree or a list, it doesn't matter. Have at it.

You are welcome to argue with logic itself, since infinite ordered logical trees are exactly logic itself.

Because you argue with logic itself, you are naturally missing http://www.internationalskeptics.com/forums/showpost.php?p=11483325&postcount=2113, which does exactly the opposite, it proves that N is (by using the standard terminology) uncountable exactly because the list is already (by using the standard terminology) uncountable.

Generally, you still try to comprehend http://www.internationalskeptics.com/forums/showpost.php?p=11481365&postcount=2101 by using Cantor's notions about infinite sets, so dear jsfisher, you are wasting you time by doing it.

 

[jsfisher]

Both sets are perfectly ordered, so there is no problem to define a bijection from each distinct N member to each distinct member of the left or right (by using the standard terminology) uncountable infinite ordered logical trees.

...and what would that ordering be?

May I assume it would start with 000...? What are the next few elements that follow?

 

[doronshadmi]

May I assume it would start with 000...? What are the next few elements that follow?

0...1
0...10
0...11
...

and it ends with 1...1 that can be (if we wish to do so) = 111...

The all cases of #...# mean infinitely many bits of the same left bit.

 

[jsfisher]

0...1
0...10
0...11
...

There are no such elements in the set.

 

[doronshadmi]

There are no such elements in the set.

There are exactly such ordered logical connectives in the left and right infinite ordered logical trees.

But infinite logical connectives that start with bit 1, are included only in the left tree.

The arbitrary mixed set has exactly quarter of the left side of the left tree AND exactly quarter of the right side of the left tree.

You have no clue what infinite ordered logical connectives are since you did not study them until this very moment.

But as long as you live it is never too late.

 

[jsfisher]

There are exactly such logical connectives in the left and right infinite ordered logical trees.

No, there are not. Infinite sequences do not come to a abrupt end somewhere out there. They just keep going. Otherwise, they wouldn't be infinite.

You are dealing with fantasy, not Mathematics. Let me know if and when you want to discuss the latter.