- #1
Organic
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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:
Therefore 2^aleph0 = aleph0
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:
Code:
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
Last edited: