Find cardinality of $X = left{ A : A subset mathbb R wedge text{c}(A) right} $












2












$begingroup$


I have some doubts with this task:




Find cardinality of a) $X = left{ A : A subset mathbb R wedge text{c}(A) right} $
b) $X = left{ A : A subset mathbb Q wedge text{c}(A) right} $
where $text{c}(A)$ means that set contains maximum and minimum element




I think that the result is $ mathfrak{c} $, so I have decided to show two injectives:
$$f:mathbb R rightarrow X $$
and
$$g:X rightarrow mathbb R $$
If it comes to $f$ it may be $$ f = lambda x.left{x right} $$ and that set contains maximum and minimum element so I think that it is good example (both in a) and b) ).

But I am trying to show example of function $g$ and I have stuck there. One idea was to take $$ g = lambda A. frac{1}{2}(min+max) $$ but it is not injective :(
thanks for your time






elementary-set-theory cardinals







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$endgroup$

















    2












    $begingroup$


    I have some doubts with this task:




    Find cardinality of a) $X = left{ A : A subset mathbb R wedge text{c}(A) right} $
    b) $X = left{ A : A subset mathbb Q wedge text{c}(A) right} $
    where $text{c}(A)$ means that set contains maximum and minimum element




    I think that the result is $ mathfrak{c} $, so I have decided to show two injectives:
    $$f:mathbb R rightarrow X $$
    and
    $$g:X rightarrow mathbb R $$
    If it comes to $f$ it may be $$ f = lambda x.left{x right} $$ and that set contains maximum and minimum element so I think that it is good example (both in a) and b) ).

    But I am trying to show example of function $g$ and I have stuck there. One idea was to take $$ g = lambda A. frac{1}{2}(min+max) $$ but it is not injective :(
    thanks for your time






    elementary-set-theory cardinals







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I have some doubts with this task:




      Find cardinality of a) $X = left{ A : A subset mathbb R wedge text{c}(A) right} $
      b) $X = left{ A : A subset mathbb Q wedge text{c}(A) right} $
      where $text{c}(A)$ means that set contains maximum and minimum element




      I think that the result is $ mathfrak{c} $, so I have decided to show two injectives:
      $$f:mathbb R rightarrow X $$
      and
      $$g:X rightarrow mathbb R $$
      If it comes to $f$ it may be $$ f = lambda x.left{x right} $$ and that set contains maximum and minimum element so I think that it is good example (both in a) and b) ).

      But I am trying to show example of function $g$ and I have stuck there. One idea was to take $$ g = lambda A. frac{1}{2}(min+max) $$ but it is not injective :(
      thanks for your time






      elementary-set-theory cardinals







      share|cite|improve this question











      $endgroup$




      I have some doubts with this task:




      Find cardinality of a) $X = left{ A : A subset mathbb R wedge text{c}(A) right} $
      b) $X = left{ A : A subset mathbb Q wedge text{c}(A) right} $
      where $text{c}(A)$ means that set contains maximum and minimum element




      I think that the result is $ mathfrak{c} $, so I have decided to show two injectives:
      $$f:mathbb R rightarrow X $$
      and
      $$g:X rightarrow mathbb R $$
      If it comes to $f$ it may be $$ f = lambda x.left{x right} $$ and that set contains maximum and minimum element so I think that it is good example (both in a) and b) ).

      But I am trying to show example of function $g$ and I have stuck there. One idea was to take $$ g = lambda A. frac{1}{2}(min+max) $$ but it is not injective :(
      thanks for your time






      elementary-set-theory cardinals




      elementary-set-theory cardinals






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      share|cite|improve this question













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      edited Jan 24 at 9:20







      VirtualUser

















      asked Jan 17 at 19:54









      VirtualUserVirtualUser

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          $begingroup$

          For (a) let $H:mathbb Rto(0,1)$ be bijective.



          Then define $f:2^{mathbb R}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so $2^{frak{c}}≤|X|$, and $X⊆ 2^{Bbb R}$ so $|X|≤2^frak c$.



          For (b) let $H:mathbb Qto mathbb Qcap (0,1)$ be bijective.



          Then define $f:2^{mathbb Q}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so ${frak c}≤|X|$, and $X⊆ 2^{Bbb Q}$ so $|X|≤frak c$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            For a), here's an injective map $mathfrak P(Bbb R)to X$:
            $$ Smapsto {,tfrac x{1+|x|}mid xin S}cup {-2,2}$$
            showing that $|X|ge 2^{mathfrak c}$ (and of course this means equality).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why $ |X|ge 2^{mathfrak c} $ ?
              $endgroup$
              – VirtualUser
              Jan 18 at 22:02



















            0












            $begingroup$

            Let $J= (-pi/2,pi/2)}$ and let $Y$ be the set of all subsets of $J.$



            Now $|J|=|Bbb R|=c.$ E.g. $f(x)=tan x$ is a bijection from $J$ to $Bbb R.$ So $|Y|={t:tsubset Bbb R}|=2^c>c.$



            For $sin Y$ let $G(s)=scup{-pi/2,pi/2}.$ Since $G$ is one-to-one and since ${G(s):sin Y}subset X,$ we have $$2^c=|Y|=|{G(s):sin Y}|le |X|le 2^c.$$ So $|X|=2^c.$






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              3 Answers
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              active

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              3 Answers
              3






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              active

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              active

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              2












              $begingroup$

              For (a) let $H:mathbb Rto(0,1)$ be bijective.



              Then define $f:2^{mathbb R}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so $2^{frak{c}}≤|X|$, and $X⊆ 2^{Bbb R}$ so $|X|≤2^frak c$.



              For (b) let $H:mathbb Qto mathbb Qcap (0,1)$ be bijective.



              Then define $f:2^{mathbb Q}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so ${frak c}≤|X|$, and $X⊆ 2^{Bbb Q}$ so $|X|≤frak c$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For (a) let $H:mathbb Rto(0,1)$ be bijective.



                Then define $f:2^{mathbb R}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so $2^{frak{c}}≤|X|$, and $X⊆ 2^{Bbb R}$ so $|X|≤2^frak c$.



                For (b) let $H:mathbb Qto mathbb Qcap (0,1)$ be bijective.



                Then define $f:2^{mathbb Q}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so ${frak c}≤|X|$, and $X⊆ 2^{Bbb Q}$ so $|X|≤frak c$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For (a) let $H:mathbb Rto(0,1)$ be bijective.



                  Then define $f:2^{mathbb R}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so $2^{frak{c}}≤|X|$, and $X⊆ 2^{Bbb R}$ so $|X|≤2^frak c$.



                  For (b) let $H:mathbb Qto mathbb Qcap (0,1)$ be bijective.



                  Then define $f:2^{mathbb Q}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so ${frak c}≤|X|$, and $X⊆ 2^{Bbb Q}$ so $|X|≤frak c$.






                  share|cite|improve this answer









                  $endgroup$



                  For (a) let $H:mathbb Rto(0,1)$ be bijective.



                  Then define $f:2^{mathbb R}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so $2^{frak{c}}≤|X|$, and $X⊆ 2^{Bbb R}$ so $|X|≤2^frak c$.



                  For (b) let $H:mathbb Qto mathbb Qcap (0,1)$ be bijective.



                  Then define $f:2^{mathbb Q}→ X$: $f(A)=H[A]∪{0,1}$, because $H$ is bijective $f$ is injective so ${frak c}≤|X|$, and $X⊆ 2^{Bbb Q}$ so $|X|≤frak c$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 24 at 11:46









                  HoloHolo

                  5,75421131




                  5,75421131























                      3












                      $begingroup$

                      For a), here's an injective map $mathfrak P(Bbb R)to X$:
                      $$ Smapsto {,tfrac x{1+|x|}mid xin S}cup {-2,2}$$
                      showing that $|X|ge 2^{mathfrak c}$ (and of course this means equality).






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Why $ |X|ge 2^{mathfrak c} $ ?
                        $endgroup$
                        – VirtualUser
                        Jan 18 at 22:02
















                      3












                      $begingroup$

                      For a), here's an injective map $mathfrak P(Bbb R)to X$:
                      $$ Smapsto {,tfrac x{1+|x|}mid xin S}cup {-2,2}$$
                      showing that $|X|ge 2^{mathfrak c}$ (and of course this means equality).






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Why $ |X|ge 2^{mathfrak c} $ ?
                        $endgroup$
                        – VirtualUser
                        Jan 18 at 22:02














                      3












                      3








                      3





                      $begingroup$

                      For a), here's an injective map $mathfrak P(Bbb R)to X$:
                      $$ Smapsto {,tfrac x{1+|x|}mid xin S}cup {-2,2}$$
                      showing that $|X|ge 2^{mathfrak c}$ (and of course this means equality).






                      share|cite|improve this answer









                      $endgroup$



                      For a), here's an injective map $mathfrak P(Bbb R)to X$:
                      $$ Smapsto {,tfrac x{1+|x|}mid xin S}cup {-2,2}$$
                      showing that $|X|ge 2^{mathfrak c}$ (and of course this means equality).







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 17 at 20:13









                      Hagen von EitzenHagen von Eitzen

                      281k23272505




                      281k23272505












                      • $begingroup$
                        Why $ |X|ge 2^{mathfrak c} $ ?
                        $endgroup$
                        – VirtualUser
                        Jan 18 at 22:02


















                      • $begingroup$
                        Why $ |X|ge 2^{mathfrak c} $ ?
                        $endgroup$
                        – VirtualUser
                        Jan 18 at 22:02
















                      $begingroup$
                      Why $ |X|ge 2^{mathfrak c} $ ?
                      $endgroup$
                      – VirtualUser
                      Jan 18 at 22:02




                      $begingroup$
                      Why $ |X|ge 2^{mathfrak c} $ ?
                      $endgroup$
                      – VirtualUser
                      Jan 18 at 22:02











                      0












                      $begingroup$

                      Let $J= (-pi/2,pi/2)}$ and let $Y$ be the set of all subsets of $J.$



                      Now $|J|=|Bbb R|=c.$ E.g. $f(x)=tan x$ is a bijection from $J$ to $Bbb R.$ So $|Y|={t:tsubset Bbb R}|=2^c>c.$



                      For $sin Y$ let $G(s)=scup{-pi/2,pi/2}.$ Since $G$ is one-to-one and since ${G(s):sin Y}subset X,$ we have $$2^c=|Y|=|{G(s):sin Y}|le |X|le 2^c.$$ So $|X|=2^c.$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Let $J= (-pi/2,pi/2)}$ and let $Y$ be the set of all subsets of $J.$



                        Now $|J|=|Bbb R|=c.$ E.g. $f(x)=tan x$ is a bijection from $J$ to $Bbb R.$ So $|Y|={t:tsubset Bbb R}|=2^c>c.$



                        For $sin Y$ let $G(s)=scup{-pi/2,pi/2}.$ Since $G$ is one-to-one and since ${G(s):sin Y}subset X,$ we have $$2^c=|Y|=|{G(s):sin Y}|le |X|le 2^c.$$ So $|X|=2^c.$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Let $J= (-pi/2,pi/2)}$ and let $Y$ be the set of all subsets of $J.$



                          Now $|J|=|Bbb R|=c.$ E.g. $f(x)=tan x$ is a bijection from $J$ to $Bbb R.$ So $|Y|={t:tsubset Bbb R}|=2^c>c.$



                          For $sin Y$ let $G(s)=scup{-pi/2,pi/2}.$ Since $G$ is one-to-one and since ${G(s):sin Y}subset X,$ we have $$2^c=|Y|=|{G(s):sin Y}|le |X|le 2^c.$$ So $|X|=2^c.$






                          share|cite|improve this answer









                          $endgroup$



                          Let $J= (-pi/2,pi/2)}$ and let $Y$ be the set of all subsets of $J.$



                          Now $|J|=|Bbb R|=c.$ E.g. $f(x)=tan x$ is a bijection from $J$ to $Bbb R.$ So $|Y|={t:tsubset Bbb R}|=2^c>c.$



                          For $sin Y$ let $G(s)=scup{-pi/2,pi/2}.$ Since $G$ is one-to-one and since ${G(s):sin Y}subset X,$ we have $$2^c=|Y|=|{G(s):sin Y}|le |X|le 2^c.$$ So $|X|=2^c.$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 24 at 21:18









                          DanielWainfleetDanielWainfleet

                          35.2k31648




                          35.2k31648






























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