Let-S-denotes-number-of-elements-in-a-set-S-N-and-R-are-sets-of-natural-and-real-numbers-respectively-N-R-

Question Number 1682 by Rasheed Soomro last updated on 31/Aug/15

L e t ∣ S ∣ d e n o t e s n u m b e r o f e l e m e n t s i n a s e t S,

N a n d R a r e s e t s o f n a t u r a l a n d r e a l n u m b e r s

r e s p e c t i v e l y :

∣ N ∣ \overset{?}{=} ∣ R ∣

Commented by 112358 last updated on 31/Aug/15

∣ N ∣\neq∣ R ∣

F o r a s e t X b e i n g a p r o p e r s u b s e t

o f a n o t h e r s e t Y, w e h a v e t h a t

∣ X ∣<∣ Y ∣ . i . e I f X \subset Y \Rightarrow∣ X ∣<∣ Y ∣ .

S i n c e N \subset R \Rightarrow∣ N ∣<∣ R ∣ . T h i s

i n f o r m a l e x p l a n a t i o n d o e s n o t

p r o v i d e a r i g o r o u s p r o o f w h i c h

s h o w s t h a t t h e i n f i n i t y o f t h e s e t

R i s l a r g e r t h a n t h a t o f t h e s e t N .

I {h a v e n}^{'} t s t u d i e d s e t t h e o r y t o a n

a d v a n c e d l e v e l . I h a v e r e a d t h a t

i f ∣ N ∣ h a s i t s v a l u e d e n o t e d b y ℵ_{0}

t h e n, a c c o r d i n g t o t h e c o n t i n u u m

h y p o t h e s i s, ℵ_{0} <∣ R ∣ .

Commented by 123456 last updated on 31/Aug/15

if two sets are equal you can make a

1 - 1 function with them, a set that

are 1 - 1 with N are called contable

however R dont is contable

you can get number into [0, 1]

0

0, 1

0, 2

\dots

and you still miss some number

in other hand Q are contable and this

is showed by diagolization argument

so ∣ Q ∣=∣ N ∣

Commented by Rasheed Ahmad last updated on 31/Aug/15

\boldsymbol X \subset \boldsymbol Y \Rightarrow∣ \boldsymbol X ∣<∣ \boldsymbol Y ∣

{1, 3, 5, \dots} \subset {1, 2, 3, \dots}

\overset{?}{\Rightarrow} ∣ {1, 3, 5, \dots} ∣ \overset{?}{<} ∣ {1, 2, 3, \dots} ∣

W h i l e :

B o t h s e t s a r e e q u i v a l e n t .

(T h e r e i s 1 - 1 c o r r e s p o n d a n c e

b e t w e e n t h e m)

T h e y a r e c o u n t a b l e a n d t h e i r

i n f i n i t i e s a r e e q u a l .

I t h i n k t h e a b o v e l a w w o r k s o n l y i n c a s e o f

f i n i t e s e t s .

Commented by 123456 last updated on 31/Aug/15

yes, if X \subset Y, ∣ X ∣⩽∣ Y ∣

try to search hilbert hotel on google

its will help yoj

in general a infinite set can also have

infinite substes that are infinites

the peoblem lead to trasnfinites number

ℵ_{0} =∣ N ∣

ℵ_{1} =∣ R ∣= 2^{ℵ_{0}}

or something like this

the continuous hipothesys s about a existence

of a set that are large than natural,

but small than real

∣ A ∣= ℵ

ℵ_{0} < ℵ < ℵ_{1}

the most strange its that the answer

is yes and no, also with set axioms

you cannot proof or disproof it

wich leads to godel imcopletude theorems

Commented by Rasheed Soomro last updated on 03/Sep/15

T^{H^{A} N} Ks a L^{O} T_{!}

T^{H^{A} N} Ks a L^{O} T_{!}