On P(X) > X

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  • #51
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the symbol 0 has many uses. 0 could be the identity element in a group.

0 could be a cardinal number or it could be a natural number. in the first case, |{}|=0 and in the second case, {}=0. it can be ambiguous that 0 is used so much. as far as i know, |{}|!={} yet people use the symbol 0 for both.

0 = |{ }| (notation = {})

1 = |{{ }}| (notation = {0})
of 1={0}, then shouldn't it be 1=||{}|| and not |{{}}|?

while this construction of N isn't necessarily "wrong," i like the ZFC better because it doesn't mention cardinality (under the general impression that less is more).
 
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  • #52
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By writing the natural numbers as ...(((0+1)+1)+1)... you clearly say that N is an hierarchic recursion.

So P(N) members are totally dependent on N members for their existence, isn’t it?
 
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  • #53
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i would say that for any set x, P(x) totally depends on x.
 
  • #54
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So, dont you think that the axiom of subsets ignore this hierarchic of dependency ?
 
  • #55
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in a way it does but it also makes a given set depend on a (bigger) set, a context.
 
  • #56
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Dear phoenixthoth,

So, we can't ignore the idea of dependency, which can be translated to the idea of the power of existence of any N memeber, which is bigger than the power of existence of any P(N) member (exclude {}, and S not= {} because {} not in N, therefore S includes the N member, which mapped with {}).

I'll write it down again, and please read it by the light of "the idea of dependency".


Please read this: http://mathworld.wolfram.com/CantorDiagonalMethod.html

N=S
P(N)=T
S=D
t=x

Now, please follow my idea until the end of what I write, and then please write your remarks.

thank you.

By using the empty set (with the Von Neumann Hierarchy), we can construct the set of all positive integers {0,1,2,3,4,...}:
Code:
[b][i]0[/i][/b] = |{ }| (notation = {}) 

[b][i]1[/i][/b] = |{[b]{[/b] [b]}[/b]}| (notation = {0})
               
[b][i]2[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b]}| (notation = {0,1})
  
[b][i]3[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b]}| (notation = {0,1,2})

[b][i]4[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b],[b]{[/b]{ },{{ }},{{ },{{ }}}[b]}[/b]}| (notation = {0,1,2,3})

and so on.
So, through this point of view the atom is {} and the natural numbers are structural|quantitative combinations of {}.

We have here an Hierarchy of infinitely many complexity levels, starting from {}.

If {} does not exist then we have no system to research, therefore we can learn that in the base of this number system, there is the idea of dependency, which can be translated to the idea of the power of existence.

Let us look at Cantor's proof from this point of view.

As we all know, any set includes only unique members or no members at all.

Now, let us examine options 1 and 2 again.

Option 1:

All members which included in S , are different from each other.

Any member of N can be mapped with some member of P(N), once and only once.

Therefore t is different from each member in S , therefore t MUST BE INCLUDED in S .

(S exists iff t is out of the scope of its definition. it means that the word ALL is omitted from S definition).

Therefore S cannot be defined, and we can't check our assumption in step 2 of Cantor's proof.

Option 2:

If we want to keep S as an existing member, we MUST NOT INCLUDE t in S .

(S exists iff t is out of the scope of its definition. it means that the word ALL is omitted from S definition).

In this case we also find that S cannot be defined, and we can't check our assumption in step 2 of Cantor's proof.


Therefore we cannot ask nor answer to anything that is connected to member S .


The general idea behind this point of view is the power of existence of member t and member S .

More you simple less you depended, therefore more exist.

Now, the power of existence of N members is bigger than the power of existence of P(N) members.

Member t, which is some arbitrary member of N(and mapped with S), is simpler than S, which is a member of P(N) .

Member S existence depends on objects like t , therefore we have to check S by t as we did here, and not t by S as Cantor did.

Conclusion: Cantor did not prove that P(N) > N because S (as Cantor defined it) does not exist.


Organic
 
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  • #57
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maybe you can use transfinite induction to prove P(N)>N. ;)

i think this boils down to a discussion of whether or not to accept the subsets axiom. there is no real right or wrong answer to that.
 
  • #58
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Proofs by transfinite induction typically need to distinguish three cases:

1) m is a minimal element, i.e. there is no element smaller than m.

2) m has a direct predecessor, i.e. the set of elements which are smaller than m has a largest element.

3) m has no direct predecessor, i.e. m is a so-called limit-ordinal.


In all cases "m does not exist" = "transfinite induction does not exist"

We have here an hierarchic of dependency that can't be ignored.

Therefore I still clime that we have to check S by t as I did, and not t by S as Cantor did.

The order of this difference (S by t XOR t by S) can't be ignored.
 
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