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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-

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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 ∣
$${Let}\:\mid\:\mathrm{S}\:\mid\:{denotes}\:{number}\:{of}\:{elements}\:{in}\:{a}\:{set}\:\mathrm{S}\:,\: \\ $$$$\mathbb{N}\:{and}\:\mathbb{R}\:{are}\:{sets}\:{of}\:{natural}\:{and}\:{real}\:{numbers}\: \\ $$$${respectively}: \\ $$$$\mid\:\mathbb{N}\:\mid\overset{?} {=}\mid\:\mathbb{R}\:\mid \\ $$
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∣.
$$\mid\mathbb{N}\mid\neq\mid\mathbb{R}\mid\: \\ $$$${For}\:{a}\:{set}\:{X}\:{being}\:{a}\:{proper}\:{subset} \\ $$$${of}\:{another}\:{set}\:{Y},\:{we}\:{have}\:{that} \\ $$$$\mid{X}\mid<\mid{Y}\mid.\:{i}.{e}\:{If}\:{X}\subset{Y}\Rightarrow\mid{X}\mid<\mid{Y}\mid. \\ $$$${Since}\:\mathbb{N}\subset\mathbb{R}\Rightarrow\mid\mathbb{N}\mid<\mid\mathbb{R}\mid.\:{This} \\ $$$${informal}\:{explanation}\:{does}\:{not} \\ $$$${provide}\:{a}\:{rigorous}\:{proof}\:{which} \\ $$$${shows}\:{that}\:{the}\:{infinity}\:{of}\:{the}\:{set} \\ $$$$\mathbb{R}\:{is}\:{larger}\:{than}\:{that}\:{of}\:{the}\:{set}\:\mathbb{N}. \\ $$$${I}\:{haven}'{t}\:{studied}\:{set}\:{theory}\:{to}\:{an} \\ $$$${advanced}\:{level}.\:{I}\:{have}\:{read}\:{that} \\ $$$${if}\:\mid\mathbb{N}\mid\:{has}\:{its}\:{value}\:{denoted}\:{by}\:\aleph_{\mathrm{0}} \\ $$$${then},{according}\:{to}\:{the}\:{continuum} \\ $$$${hypothesis},\aleph_{\mathrm{0}} <\mid\mathbb{R}\mid.\: \\ $$$$ \\ $$$$ \\ $$
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∣
$$\mathrm{if}\:\mathrm{two}\:\mathrm{sets}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{you}\:\mathrm{can}\:\mathrm{make}\:\mathrm{a} \\ $$$$\mathrm{1}−\mathrm{1}\:\mathrm{function}\:\mathrm{with}\:\mathrm{them},\:\mathrm{a}\:\mathrm{set}\:\mathrm{that} \\ $$$$\mathrm{are}\:\mathrm{1}−\mathrm{1}\:\mathrm{with}\:\mathbb{N}\:\mathrm{are}\:\mathrm{called}\:\mathrm{contable} \\ $$$$\mathrm{however}\:\mathbb{R}\:\mathrm{dont}\:\mathrm{is}\:\mathrm{contable} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{get}\:\mathrm{number}\:\mathrm{into}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{0} \\ $$$$\mathrm{0},\mathrm{1} \\ $$$$\mathrm{0},\mathrm{2} \\ $$$$… \\ $$$$\mathrm{and}\:\mathrm{you}\:\mathrm{still}\:\mathrm{miss}\:\mathrm{some}\:\mathrm{number} \\ $$$$\mathrm{in}\:\mathrm{other}\:\mathrm{hand}\:\mathbb{Q}\:\mathrm{are}\:\mathrm{contable}\:\mathrm{and}\:\mathrm{this} \\ $$$$\mathrm{is}\:\mathrm{showed}\:\mathrm{by}\:\mathrm{diagolization}\:\mathrm{argument} \\ $$$$\mathrm{so}\:\mid\mathbb{Q}\mid=\mid\mathbb{N}\mid \\ $$
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.
$$\boldsymbol{\mathrm{X}}\subset\boldsymbol{\mathrm{Y}}\:\Rightarrow\mid\boldsymbol{\mathrm{X}}\mid<\mid\boldsymbol{\mathrm{Y}}\mid \\ $$$$\left\{\mathrm{1},\mathrm{3},\mathrm{5},…\right\}\subset\left\{\mathrm{1},\mathrm{2},\mathrm{3},…\right\} \\ $$$$\overset{?} {\Rightarrow}\mid\:\left\{\mathrm{1},\mathrm{3},\mathrm{5},…\right\}\:\mid\:\overset{?} {<}\:\mid\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},…\right\}\mid \\ $$$${While}: \\ $$$${Both}\:{sets}\:{are}\:{equivalent}. \\ $$$$\left({There}\:{is}\:\mathrm{1}−\mathrm{1}\:{correspondance}\right. \\ $$$$\left.{between}\:{them}\right) \\ $$$${They}\:{are}\:{countable}\:{and}\:{their} \\ $$$${infinities}\:{are}\:{equal}. \\ $$$${I}\:{think}\:{the}\:{above}\:{law}\:{works}\:{only}\:{in}\:{case}\:{of} \\ $$$${finite}\:{sets}. \\ $$$$ \\ $$
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
$$\mathrm{yes},\:\mathrm{if}\:\mathrm{X}\subset\mathrm{Y},\mid\mathrm{X}\mid\leqslant\mid\mathrm{Y}\mid \\ $$$$\mathrm{try}\:\mathrm{to}\:\mathrm{search}\:\mathrm{hilbert}\:\mathrm{hotel}\:\mathrm{on}\:\mathrm{google} \\ $$$$\mathrm{its}\:\mathrm{will}\:\mathrm{help}\:\mathrm{yoj} \\ $$$$\mathrm{in}\:\mathrm{general}\:\mathrm{a}\:\mathrm{infinite}\:\mathrm{set}\:\mathrm{can}\:\mathrm{also}\:\mathrm{have} \\ $$$$\mathrm{infinite}\:\mathrm{substes}\:\mathrm{that}\:\mathrm{are}\:\mathrm{infinites} \\ $$$$\mathrm{the}\:\mathrm{peoblem}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{trasnfinites}\:\mathrm{number} \\ $$$$\aleph_{\mathrm{0}} =\mid\mathbb{N}\mid \\ $$$$\aleph_{\mathrm{1}} =\mid\mathbb{R}\mid=\mathrm{2}^{\aleph_{\mathrm{0}} } \\ $$$$\mathrm{or}\:\mathrm{something}\:\mathrm{like}\:\mathrm{this} \\ $$$$\mathrm{the}\:\mathrm{continuous}\:\mathrm{hipothesys}\:\mathrm{s}\:\mathrm{about}\:\mathrm{a}\:\mathrm{existence} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{set}\:\mathrm{that}\:\mathrm{are}\:\mathrm{large}\:\mathrm{than}\:\mathrm{natural}, \\ $$$$\mathrm{but}\:\mathrm{small}\:\mathrm{than}\:\mathrm{real} \\ $$$$\mid\mathrm{A}\mid=\aleph \\ $$$$\aleph_{\mathrm{0}} <\aleph<\aleph_{\mathrm{1}} \\ $$$$\mathrm{the}\:\mathrm{most}\:\mathrm{strange}\:\mathrm{its}\:\mathrm{that}\:\mathrm{the}\:\mathrm{answer} \\ $$$$\mathrm{is}\:\mathrm{yes}\:\mathrm{and}\:\mathrm{no},\:\mathrm{also}\:\mathrm{with}\:\mathrm{set}\:\mathrm{axioms} \\ $$$$\mathrm{you}\:\mathrm{cannot}\:\mathrm{proof}\:\mathrm{or}\:\mathrm{disproof}\:\mathrm{it} \\ $$$$\mathrm{wich}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{godel}\:\mathrm{imcopletude}\:\mathrm{theorems} \\ $$
Commented by Rasheed Soomro last updated on 03/Sep/15
T^(H^A N) Ks a L^O T_!
$$\mathrm{T}^{\mathrm{H}^{\mathrm{A}} \mathrm{N}} \mathrm{Ks}\:\mathrm{a}\:\mathrm{L}^{\mathrm{O}} \mathrm{T}_{!} \: \\ $$

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