Let ∣ S ∣ denotes number of elements in a set S , N and R are sets of natural and real numbers respectively: ∣ N ∣=^(?) ∣ R ∣


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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

$if two sets are equal you can make a$ $1 - 1 function with them, a set that$ $are 1 - 1 with N are called contable$ $however R dont is contable$ $you can get number into [0, 1]$ $0$ $0, 1$ $0, 2$ $. . .$ $and you still miss some number$ $in other hand Q are contable and this$ $is showed by diagolization argument$ $so ∣ Q ∣=∣ N ∣$
Commented by Rasheed Ahmad last updated on 31/Aug/15
$X⊂Y ⇒∣X∣<∣Y∣ {1,3,5,...}⊂{1,2,3,...} ⇒^? ∣ {1,3,5,...} ∣ <^(?) ∣ {1,2,3,...}∣ While: Both sets are equivalent. (There is 1−1 correspondance between them) They are countable and their infinities are equal. I think the above law works only in case of finite sets.$
$X \subset Y \Rightarrow∣ X ∣<∣ Y ∣$ ${1, 3, 5, . . .} \subset {1, 2, 3, . . .}$ $\overset{?}{\Rightarrow} ∣ {1, 3, 5, . . .} ∣ \overset{?}{<} ∣ {1, 2, 3, . . .} ∣$ $W h i l e :$ $B o t h s e t s a r e e q u i v a l e n t .$ $(T h e r e i s 1 - 1 c o r r e s p o n d a n c e$ $b e t w e e n t h e m)$ $T h e y a r e c o u n t a b l e a n d t h e i r$ $i n f i n i t i e s a r e e q u a l .$ $I t h i n k t h e a b o v e l a w w o r k s o n l y i n c a s e o f$ $f i n i t e s e t s .$
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

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