2^aleph0 < c

[Doron Shadmi]

Hi ,




The question is:


Theorem: 2^aleph0 < c


Proof:

Let A be the set of all negative real numbers included in (-1,0).

Let B be the set of all positive real numbers included in (0,1).

Let M be the set of maps (1 to 1 and onto) between any two single numbers of A and B sets.

Therefore |M| = 2^aleph0.

(0,1) = {x: 0 < x < 1 }, where x is a 1-1 correspondence between any two real numbers included in (0,1),
and any x element has no more than 1 real number as a common element with some other x element.

Let T be the set of all x (1-1 correspondence) elements included in (0,1).

Therefore |T| = |M| = 2^aleph0.

B is a totally ordered set, therefore we can find x element between any two
different real numbers included in (0,1).

Any x element must be > 0 and cannot include in it any real number.

Therefore 2^aleph0 < c (does not have the power of the continuum).

Q.E.D

A structural model of the above:

            set      set       set  
             A        M         B
             |        |         |
             |        |         |
             v        v         v
             !__________________!<---- set
             !__________________!<----  T members 
             !__________________!
             !__________________!   Any point is some real number 
             !__________________!   (A or B members).
             !__________________!   
             !__________________!   Any line is some 1-1 correspondence    
             !__________________!   (M or T members).
             !__________________!
             !__________________!

Am I right ?