How to show $2^{ℵ_0} leq mathfrak c$ [duplicate]












6












$begingroup$



This question already has an answer here:




  • Easiest way to prove that $2^{aleph_0} = c$

    3 answers




I want to show $2^{ℵ_0}=mathfrak c$.



I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:



Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have



$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$



But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?






elementary-set-theory cardinals







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 Dec 11 '18 at 22:59


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    Look at the Cantor set.
    $endgroup$
    – Andrés E. Caicedo
    Nov 26 '18 at 13:04
















6












$begingroup$



This question already has an answer here:




  • Easiest way to prove that $2^{aleph_0} = c$

    3 answers




I want to show $2^{ℵ_0}=mathfrak c$.



I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:



Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have



$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$



But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$



marked as duplicate by Lord_Farin, amWhy elementary-set-theory
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 Dec 11 '18 at 22:59


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    Look at the Cantor set.
    $endgroup$
    – Andrés E. Caicedo
    Nov 26 '18 at 13:04














6












6








6


1



$begingroup$



This question already has an answer here:




  • Easiest way to prove that $2^{aleph_0} = c$

    3 answers




I want to show $2^{ℵ_0}=mathfrak c$.



I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:



Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have



$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$



But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?






elementary-set-theory cardinals







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Easiest way to prove that $2^{aleph_0} = c$

    3 answers




I want to show $2^{ℵ_0}=mathfrak c$.



I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:



Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have



$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$



But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?





This question already has an answer here:




  • Easiest way to prove that $2^{aleph_0} = c$

    3 answers







elementary-set-theory cardinals




elementary-set-theory cardinals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 13:14









user126154

5,378716




5,378716










asked Nov 26 '18 at 12:49









user8314628user8314628

1817




1817




marked as duplicate by Lord_Farin, amWhy elementary-set-theory
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 Dec 11 '18 at 22:59


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Lord_Farin, amWhy elementary-set-theory
Users with the  elementary-set-theory badge can single-handedly close elementary-set-theory questions as duplicates and reopen them as needed.

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 Dec 11 '18 at 22:59


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Look at the Cantor set.
    $endgroup$
    – Andrés E. Caicedo
    Nov 26 '18 at 13:04














  • 1




    $begingroup$
    Look at the Cantor set.
    $endgroup$
    – Andrés E. Caicedo
    Nov 26 '18 at 13:04








1




1




$begingroup$
Look at the Cantor set.
$endgroup$
– Andrés E. Caicedo
Nov 26 '18 at 13:04




$begingroup$
Look at the Cantor set.
$endgroup$
– Andrés E. Caicedo
Nov 26 '18 at 13:04










2 Answers
2






active

oldest

votes


















1












$begingroup$

$2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$

    Then this function is injective. So $$
    text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
    leq text{Cardinal}(mathbb{R}) $$     So $2^{ℵ_0} leq mathfrak c$.






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
      $endgroup$
      – Brahadeesh
      Nov 26 '18 at 13:28










    • $begingroup$
      No. It's One to One and your saying is false%%
      $endgroup$
      – Darmad
      Nov 26 '18 at 13:32












    • $begingroup$
      Well, I'll let you know that the downvote is mine.
      $endgroup$
      – Brahadeesh
      Nov 26 '18 at 13:40










    • $begingroup$
      But why! was my solution false?
      $endgroup$
      – Darmad
      Nov 26 '18 at 13:50






    • 1




      $begingroup$
      @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
      $endgroup$
      – Brahadeesh
      Nov 27 '18 at 15:36


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.






        share|cite|improve this answer









        $endgroup$



        $2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 '18 at 13:09









        user126154user126154

        5,378716




        5,378716























            1












            $begingroup$

            You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$

            Then this function is injective. So $$
            text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
            leq text{Cardinal}(mathbb{R}) $$             So $2^{ℵ_0} leq mathfrak c$.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:28










            • $begingroup$
              No. It's One to One and your saying is false%%
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:32












            • $begingroup$
              Well, I'll let you know that the downvote is mine.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:40










            • $begingroup$
              But why! was my solution false?
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:50






            • 1




              $begingroup$
              @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
              $endgroup$
              – Brahadeesh
              Nov 27 '18 at 15:36
















            1












            $begingroup$

            You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$

            Then this function is injective. So $$
            text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
            leq text{Cardinal}(mathbb{R}) $$             So $2^{ℵ_0} leq mathfrak c$.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:28










            • $begingroup$
              No. It's One to One and your saying is false%%
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:32












            • $begingroup$
              Well, I'll let you know that the downvote is mine.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:40










            • $begingroup$
              But why! was my solution false?
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:50






            • 1




              $begingroup$
              @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
              $endgroup$
              – Brahadeesh
              Nov 27 '18 at 15:36














            1












            1








            1





            $begingroup$

            You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$

            Then this function is injective. So $$
            text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
            leq text{Cardinal}(mathbb{R}) $$             So $2^{ℵ_0} leq mathfrak c$.






            share|cite|improve this answer











            $endgroup$



            You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$

            Then this function is injective. So $$
            text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
            leq text{Cardinal}(mathbb{R}) $$             So $2^{ℵ_0} leq mathfrak c$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 26 '18 at 14:16

























            answered Nov 26 '18 at 13:23









            DarmadDarmad

            538112




            538112








            • 2




              $begingroup$
              It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:28










            • $begingroup$
              No. It's One to One and your saying is false%%
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:32












            • $begingroup$
              Well, I'll let you know that the downvote is mine.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:40










            • $begingroup$
              But why! was my solution false?
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:50






            • 1




              $begingroup$
              @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
              $endgroup$
              – Brahadeesh
              Nov 27 '18 at 15:36














            • 2




              $begingroup$
              It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:28










            • $begingroup$
              No. It's One to One and your saying is false%%
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:32












            • $begingroup$
              Well, I'll let you know that the downvote is mine.
              $endgroup$
              – Brahadeesh
              Nov 26 '18 at 13:40










            • $begingroup$
              But why! was my solution false?
              $endgroup$
              – Darmad
              Nov 26 '18 at 13:50






            • 1




              $begingroup$
              @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
              $endgroup$
              – Brahadeesh
              Nov 27 '18 at 15:36








            2




            2




            $begingroup$
            It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
            $endgroup$
            – Brahadeesh
            Nov 26 '18 at 13:28




            $begingroup$
            It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
            $endgroup$
            – Brahadeesh
            Nov 26 '18 at 13:28












            $begingroup$
            No. It's One to One and your saying is false%%
            $endgroup$
            – Darmad
            Nov 26 '18 at 13:32






            $begingroup$
            No. It's One to One and your saying is false%%
            $endgroup$
            – Darmad
            Nov 26 '18 at 13:32














            $begingroup$
            Well, I'll let you know that the downvote is mine.
            $endgroup$
            – Brahadeesh
            Nov 26 '18 at 13:40




            $begingroup$
            Well, I'll let you know that the downvote is mine.
            $endgroup$
            – Brahadeesh
            Nov 26 '18 at 13:40












            $begingroup$
            But why! was my solution false?
            $endgroup$
            – Darmad
            Nov 26 '18 at 13:50




            $begingroup$
            But why! was my solution false?
            $endgroup$
            – Darmad
            Nov 26 '18 at 13:50




            1




            1




            $begingroup$
            @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
            $endgroup$
            – Brahadeesh
            Nov 27 '18 at 15:36




            $begingroup$
            @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
            $endgroup$
            – Brahadeesh
            Nov 27 '18 at 15:36



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