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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
Let ∣ S ∣ denotes number of elements in a set S ,   N and R are sets of natural and real numbers   respectively:  ∣ N ∣=^(?) ∣ R ∣
LetSdenotesnumberofelementsinasetS,NandRaresetsofnaturalandrealnumbersrespectively:N=?R
Commented by 112358 last updated on 31/Aug/15
∣N∣≠∣R∣   For a set X being a proper subset  of another set Y, we have that  ∣X∣<∣Y∣. i.e If X⊂Y⇒∣X∣<∣Y∣.  Since N⊂R⇒∣N∣<∣R∣. This  informal explanation does not  provide a rigorous proof which  shows that the infinity of the set  R is larger than that of the set N.  I haven′t studied set theory to an  advanced level. I have read that  if ∣N∣ has its value denoted by ℵ_0   then,according to the continuum  hypothesis,ℵ_0 <∣R∣.
N∣≠∣RForasetXbeingapropersubsetofanothersetY,wehavethatX∣<∣Y.i.eIfXY⇒∣X∣<∣Y.SinceNR⇒∣N∣<∣R.ThisinformalexplanationdoesnotprovidearigorousproofwhichshowsthattheinfinityofthesetRislargerthanthatofthesetN.Ihaventstudiedsettheorytoanadvancedlevel.IhavereadthatifNhasitsvaluedenotedby0then,accordingtothecontinuumhypothesis,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∣
iftwosetsareequalyoucanmakea11functionwiththem,asetthatare11withNarecalledcontablehoweverRdontiscontableyoucangetnumberinto[0,1]00,10,2andyoustillmisssomenumberinotherhandQarecontableandthisisshowedbydiagolizationargumentsoQ∣=∣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.
XY⇒∣X∣<∣Y{1,3,5,}{1,2,3,}?{1,3,5,}<?{1,2,3,}While:Bothsetsareequivalent.(Thereis11correspondancebetweenthem)Theyarecountableandtheirinfinitiesareequal.Ithinktheabovelawworksonlyincaseoffinitesets.
Commented by 123456 last updated on 31/Aug/15
yes, if X⊂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
yes,ifXY,X∣⩽∣Ytrytosearchhilberthotelongoogleitswillhelpyojingeneralainfinitesetcanalsohaveinfinitesubstesthatareinfinitesthepeoblemleadtotrasnfinitesnumber0=∣N1=∣R∣=20orsomethinglikethisthecontinuoushipothesyssaboutaexistenceofasetthatarelargethannatural,butsmallthanrealA∣=0<<1themoststrangeitsthattheanswerisyesandno,alsowithsetaxiomsyoucannotproofordisproofitwichleadstogodelimcopletudetheorems
Commented by Rasheed Soomro last updated on 03/Sep/15
T^(H^A N) Ks a L^O T_!
THANKsaLOT!

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