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: Code: 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 ?
