Cont: Deeper than primes - Continuation 2

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doronshadmi said:
The infinite binary tree has ℵ0 levels, such that any cardinal number (represented by positional notation, where the radix point is placed at the ℵ0 level) has its own unique path of ℵ0 0;1 bits in this tree (every number is distinguished of the rest of the numbers).
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There is no such level.
 
abaddon said:
ℵ0 is defined as the set of people who give these notions any credence. It is a null set.
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The definition you quoted is standard. It was the later discussion that went off the rails.

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jsfisher said:
There is no such level.
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It is the smallest infinite level of The Infinite Binary Tree, which is greater than any finite level of that tree.

If one denies it, one actually denies the smallest infinite cardinal, which is greater than any finite cardinal (known as ℵ0).
 
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doronshadmi said:
It is the smallest infinite level of The Infinite Binary Tree, which is greater than any finite level of that tree.

If one denies it, one actually denies the smallest infinite cardinal, which is greater than any finite cardinal (known as ℵ0).
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You assume incorrectly. There is no last level to consider as your place for the radix point since every level has a level below it.

You've bashed against this mathematical reality before. It does not yield to your miscomprehension.


ETA: By the way, all of the levels of your infinite complete binary tree occur at a "finite level". There is no "infinite level", smallest or otherwise. The levels are naturally ordered and they can be labeled using the ordered set of natural numbers. No level would be labeled ℵ0.
 
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jsfisher said:
You assume incorrectly. There is no last level to consider as your place for the radix point since every level has a level below it.

You've bashed against this mathematical reality before. It does not yield to your miscomprehension.


ETA: By the way, all of the levels of your infinite complete binary tree occur at a "finite level". There is no "infinite level", smallest or otherwise. The levels are naturally ordered and they can be labeled using the ordered set of natural numbers. No level would be labeled ℵ0.
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The Infinite Binary Tree is infinite exactly because it is an endless growing whole, such that the complementary relationship between its left and it right sides is invariant during its endless growth.

The smallest infinite level of The Infinite Binary Tree, which is greater than any finite level of that tree, is notated by ℵ0.

The next infinite level is notated by ℵ1 etc., yet it has also an invariant property exactly because it is an endless growing whole, such that the complementary relationship between its left and it right sides is invariant during its endless growth.

Whole things are both variant AND invariant (as given in the examples in http://www.internationalskeptics.com/forums/showpost.php?p=12008521&postcount=2740) without getting into contradiction (as complete things do exactly because they are invariant-only).

As long as you get The Infinite Binary Tree in terms of completeness instead of in terms of wholeness, you are missing the issue at hand.

jsfisher said:
... levels are naturally ordered and they can be labeled using the ordered set of natural numbers.
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jsfisher said:
There is no last level to consider as your place for the radix point since every level has a level below it.
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n levels are finitely labeled. This is not the case with ℵ0, ℵ1 etc.
 
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An example: 111...111. or 111...110. (infinite levels) are not the same as 111111. or 111110. (finite levels).

Moreover, infinite levels can have finite values, for example: 000...111. or 000...110.

000...000. < 000...001. < 000...010. < 000...011. ... < ... 111...100. < 111...101. < 111...110. < 111...111. such that the radix point is at ℵ0, ℵ1 etc.
 
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By this notion the notation ℵ-1 defines the domain of finite levels, such that ℵ-1 < ℵ0 < ℵ1 etc.
 
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jsfisher said:
Try as you may, you cannot count your way to infinity. There is no such level.
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I do not have to count my way to infinity, since ℵ0 is the smallest level at infinity in The Infinite Binary Tree. Without it, this tree can't be defined as an infinite mathematical object, in the first place.
 
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doronshadmi said:
I do not have to count my way to infinity, since ℵ0 is the smallest level at infinity in The Infinite Binary Tree. Without it, this tree can't be defined as an infinite mathematical object, in the first place.
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It is not an infinite mathematical object. It is an object with an infinite number of levels. Countably infinite, in fact, just like the whole numbers are countably infinite.

There is no ℵ0 in the set of whole numbers; there is no level corresponding to ℵ0 in your infinite, complete binary tree.
 
jsfisher said:
It is an object with an infinite number of levels.
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You are right, and the radix point is at least at the level that is greater than any finite level of The Infinite Binary Tree.

jsfisher said:
complete binary tree.
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This is exactly where you fail, because the tree is an infinite whole (variant AND invariant) and not infinite complete (invariant-only), as already shown in http://www.internationalskeptics.com/forums/showpost.php?p=12008521&postcount=2740, http://www.internationalskeptics.com/forums/showpost.php?p=12009500&postcount=2746, http://www.internationalskeptics.com/forums/showpost.php?p=12009549&postcount=2747 and http://www.internationalskeptics.com/forums/showpost.php?p=12009600&postcount=2748.
 
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doronshadmi said:
You are right, and the radix point is at least at the level that is greater than any finite level of The Infinite Binary Tree.
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Every level of the tree is at a finite level, just like every member of the set of whole numbers is a finite number.

This is exactly where you fail, because the tree is an infinite whole....
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You might want to look up the meaning of "complete binary tree". You have gone off on an irrelevant tangent.
 
jsfisher said:
Every level of the tree is at a finite level, just like every member of the set of whole numbers is a finite number.
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Yet the smallest infinite level of that tree is at a level which is greater than every finite level of that tree.

You see jsfisher, I am using natural inclusion AND exclusion of that tree (looking at that tree from within and from without as a whole (variant AND invariant) thing), where you are using only natural exclusion (looking at that tree only from without as a complete (invariant-only) thing).


jsfisher said:
You might want to look up the meaning of "complete binary tree". You have gone off on an irrelevant tangent.
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You might want to deduce (by using both your visual_spatial AND verbal_symbolic brain skills) that The Infinite Binary Tree is an infinite whole (variant AND invariant without getting into contradiction, as given in the links of http://www.internationalskeptics.com/forums/showpost.php?p=12009858&postcount=2753) and not an infinite complete (invariant-only) as given by the traditional school of deduction, which its notions and results are derived from using only the verba_symbolic brain skills.
 
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doronshadmi said:
Yet the smallest infinite level of that tree is at a level which is greater than every finite level of that tree.
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There is no infinite level in the tree.

Every level can be matched to a natural number in a bijective arrangement. There can be no infinite level as that would require a level with no corresponding natural number.
 
jsfisher said:
There is no infinite level in the tree.

Every level can be matched to a natural number in a bijective arrangement. There can be no infinite level as that would require a level with no corresponding natural number.
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Once again, you exclude the smallest infinite level of that tree as if it exists externaly w.r.t it, and as a result every level of that tree is deduced by you in terms of finitism.

But if you deduce this tree both internally AND externally, levels like ℵ0, ℵ1, ℵ2 etc. are included in that tree AND also there is always an external room for further growth.

Such deduction is possible only if both visual_spatial AND verbal_symbolic brain skills are used.

since you are using only your verbal_symbolic brain skills during deduction, you can't comprehend, for example, http://www.internationalskeptics.com/forums/showpost.php?p=12009549&postcount=2747.
 
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doronshadmi said:
...But if you deduce this tree both internally AND externally...
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Your own special view of Mathematics once again gets you the wrong answer.

And once again, I will remind you that you don't get to redefine Mathematics to accommodate your misunderstandings. You binary tree has countably infinite levels. None of them correspond to an infinite level. These facts do not change just because you may disagree.
 
doronshadmi said:
Once again, you exclude the smallest infinite level of that tree as if it exists externaly w.r.t it, and as a result every level of that tree is deduced by you in terms of finitism.
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I exclude nothing of the tree that actually exists. You imagine things that are not there.

Is it your contention that the levels of your tree do not have a bijection with the natural numbers?
 
Please observe 111...111. or 111111.

The first is infinite, the second is finite but in both cases the radix point is not one of the bits, yet it is included in the tree, which means that it is a level of the tree.

This level may be finite or infinite according to the number of the bits, which can be finite or infinite.

111...111. is endless, such that the radix point level can be in any domain, finite or infinite.

By this notion the notation ℵ-1 defines the domain of finite number of the bits, ℵ0 is the first infinite domain above ℵ-1, ℵ1 is the second infinite domain above ℵ-1 etc. such that ℵ-1 < ℵ0 < ℵ1 etc.

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jsfisher said:
Is it your contention that the levels of your tree do not have a bijection with the natural numbers?
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A bijection with the natural numbers is true in case of ℵ-1
 
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