How can we represent the infinite powerset of a set?"

[Organic]

Let us check these lists.


P(2) = {{},{0},{1},{0,1}} = 2^2 = 4

and also can be represented as:

00
01
10
11


P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8

and also can be represented as:

000
001
010
011
100
101
110
111


...


P(aleph0) = 2^aleph0 = |R|

and also can be represented as:

aleph0
^
|
|
0(...--> aleph0)0
0(...--> aleph0)1
0(...--> aleph0)0
0(...--> aleph0)1
1(...--> aleph0)0
1(...--> aleph0)1
1(...--> aleph0)0
1(...--> aleph0)1
|
|
v
aleph0

We can find a bijection between N and R by this way:

aleph0
^
|
|
7 <--> 0.0(...--> aleph0)0
5 <--> 0.0(...--> aleph0)1
3 <--> 0.0(...--> aleph0)0
1 <--> 0.0(...--> aleph0)1
2 <--> 0.1(...--> aleph0)0
4 <--> 0.1(...--> aleph0)1
6 <--> 0.1(...--> aleph0)0
8 <--> 0.1(...--> aleph0)1
|
|
v
aleph0

Therefore 2^aleph0 = aleph0

 

[Organic]

But there is another thing that i have found.

We still be able to use Cantor's function and get some number which is not in the list.

For example:

aleph0
^
|
|
0.0,x23,x33,x43,... ,0
0.0,x21,x31,x41,... ,1
0.1,x22,x32,x42,... ,0
0.1,x24,x34,x44,... ,1
|
|
v
aleph0

Our new result, which is not in the list, is the opposite of 0.0,x22,x33,x44,...

So in this point we are maybe in a logical disaster.

I think the sulotion is to use the idea of the open interval on a single number.

For example:

[0.x1,x2,x3,x4,... ,1)

More information you can find here:

http://www.geocities.com/complementarytheory/RiemannsBall.pdf

 

[MathematicalPhysicist]

Originally posted by Organic
Let us check these lists.


P(2) = {{},{0},{1},{0,1}} = 2^2 = 4

I am not an expert in set theory but doesn't {}={0}=0 and so it would be useless to write it in different ways?

 

[Organic]

Hi loop quantum gravity,

{{}} = {0}

 

[Hurkyl]

\{ \} \neq \{ 0 \}. In general, 0 \neq \{ \}, but some models (including the one typically used in set theory) do make that identification.


Organic: you're missing a very important fact about the ordering structure of the integers:

If a sequence of integers has a first element and a last element, then the sequence is finite.

By definition, the digits in a decimal expansion are indexed by integers...

This means that:

0.0,x23,x33,x43,... ,0

(which I'm assuming is supposed to be some real number written in b-ary^1[/size] notation) cannot have an infinite number of digits.

Similarly

0(...--> aleph0)0

cannot be a sequence^2[/size] of binary digits.

(by this notation I'm assuming you mean that there are countably infinite numbers between the first 0 and the last 0)

Also,

7 <--> 0.0(...--> aleph0)0

cannot be a binary expansion of a real number.

(again I'm assuming that this notation means there are countably infinite numbers between the first and last digit)

Furthermore

The notation P(A) requires A to be a set. You should be saying P(\{ 0, 1\}) instead of P(2), and you should definitely be saying P(\mathbb{N}) instead of P(\aleph_0). (That is, assuming I understand correctly what you mean)

Moreover

the way you are listing the elements of the powerset of a finite set does not generalize to an infinite set. In particular, every element of the list:

...0000
...0001
...0010
...0011
...0100
...

has a finite number of ones. This is only a representation of the finite subsets of \mathbb{N}.


Additionally

I think the sulotion is to use the idea of the open interval on a single number.

For example:

[0.x1,x2,x3,x4,... ,1)

This is NOT the idea of an open interval. You should tell us what this means or stop using it.


footnotes:

1: b-ary means base-b representation. e.g. binary is 2-ary, decimal is 10-ary

2: Unless otherwise specified, a sequence is indexed by some segment the natural numbers.