How do the authors prove that “The relative complement of the Cantor set in $[0,1]$ is dense in $[0,1]$”?












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In my textbook Introduction to Set Theory by Hrbacek and Jech, the authors first construct Cantor set:




enter image description here



enter image description here



enter image description here




Next they prove The relative complement of the Cantor set in $[0,1]$ is dense in $[0,1]$:




enter image description here




My question: I can not understand why the authors conclude




The open interval $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is certainly disjoint from $F$.




Could you please elaborate on this statement?



Thank you so much!






proof-explanation cantor-set







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    In my textbook Introduction to Set Theory by Hrbacek and Jech, the authors first construct Cantor set:




    enter image description here



    enter image description here



    enter image description here




    Next they prove The relative complement of the Cantor set in $[0,1]$ is dense in $[0,1]$:




    enter image description here




    My question: I can not understand why the authors conclude




    The open interval $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is certainly disjoint from $F$.




    Could you please elaborate on this statement?



    Thank you so much!






    proof-explanation cantor-set







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      In my textbook Introduction to Set Theory by Hrbacek and Jech, the authors first construct Cantor set:




      enter image description here



      enter image description here



      enter image description here




      Next they prove The relative complement of the Cantor set in $[0,1]$ is dense in $[0,1]$:




      enter image description here




      My question: I can not understand why the authors conclude




      The open interval $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is certainly disjoint from $F$.




      Could you please elaborate on this statement?



      Thank you so much!






      proof-explanation cantor-set







      share|cite|improve this question









      $endgroup$




      In my textbook Introduction to Set Theory by Hrbacek and Jech, the authors first construct Cantor set:




      enter image description here



      enter image description here



      enter image description here




      Next they prove The relative complement of the Cantor set in $[0,1]$ is dense in $[0,1]$:




      enter image description here




      My question: I can not understand why the authors conclude




      The open interval $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is certainly disjoint from $F$.




      Could you please elaborate on this statement?



      Thank you so much!






      proof-explanation cantor-set




      proof-explanation cantor-set






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 19 at 14:24









      Le Anh DungLe Anh Dung

      1,1831621




      1,1831621






















          1 Answer
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          $begingroup$

          We remove the middle third of the interval to define the next $F_n$ and so it will be disjoint from $F_{n+1}$ so a fortiori from $F$.



          Explanation upon request: An interval from $F_n$ is of the form $[frac{k}{3^n}, frac{k+1}{3^n}]$ (not all such intervals are in $F_n$ but $2^n$ of them are; this makes the exact formula for $F-n$ tricky: specify which $k$ do occur; hence the recursive definition with sequences etc.) and we can also write this as $[frac{3k}{3^{n+1}}, frac{3(k+1)}{3^{n+1}}] = [frac{3k}{3^{n+1}}, frac{3k+3)}{3^{n+1}}]$, multiplying both parts of the fraction by $3$ and so its middle third open interval is $(frac{3k+1}{3^{n+1}}, frac{3k+2)}{3^{n+1}})$, exactly as claimed.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            From the definition of $F$, I am unable to infer that $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is the middle third of some interval from $F_n$ for some $n$. Please elaborate more!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 14:47












          • $begingroup$
            @LeAnhDung added some explanation.
            $endgroup$
            – Henno Brandsma
            Jan 19 at 14:56










          • $begingroup$
            Thank you so much! Your explanation is straight to my confusion and thus amazing!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 15:35











          Your Answer





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          1 Answer
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          active

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          1












          $begingroup$

          We remove the middle third of the interval to define the next $F_n$ and so it will be disjoint from $F_{n+1}$ so a fortiori from $F$.



          Explanation upon request: An interval from $F_n$ is of the form $[frac{k}{3^n}, frac{k+1}{3^n}]$ (not all such intervals are in $F_n$ but $2^n$ of them are; this makes the exact formula for $F-n$ tricky: specify which $k$ do occur; hence the recursive definition with sequences etc.) and we can also write this as $[frac{3k}{3^{n+1}}, frac{3(k+1)}{3^{n+1}}] = [frac{3k}{3^{n+1}}, frac{3k+3)}{3^{n+1}}]$, multiplying both parts of the fraction by $3$ and so its middle third open interval is $(frac{3k+1}{3^{n+1}}, frac{3k+2)}{3^{n+1}})$, exactly as claimed.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            From the definition of $F$, I am unable to infer that $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is the middle third of some interval from $F_n$ for some $n$. Please elaborate more!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 14:47












          • $begingroup$
            @LeAnhDung added some explanation.
            $endgroup$
            – Henno Brandsma
            Jan 19 at 14:56










          • $begingroup$
            Thank you so much! Your explanation is straight to my confusion and thus amazing!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 15:35
















          1












          $begingroup$

          We remove the middle third of the interval to define the next $F_n$ and so it will be disjoint from $F_{n+1}$ so a fortiori from $F$.



          Explanation upon request: An interval from $F_n$ is of the form $[frac{k}{3^n}, frac{k+1}{3^n}]$ (not all such intervals are in $F_n$ but $2^n$ of them are; this makes the exact formula for $F-n$ tricky: specify which $k$ do occur; hence the recursive definition with sequences etc.) and we can also write this as $[frac{3k}{3^{n+1}}, frac{3(k+1)}{3^{n+1}}] = [frac{3k}{3^{n+1}}, frac{3k+3)}{3^{n+1}}]$, multiplying both parts of the fraction by $3$ and so its middle third open interval is $(frac{3k+1}{3^{n+1}}, frac{3k+2)}{3^{n+1}})$, exactly as claimed.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            From the definition of $F$, I am unable to infer that $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is the middle third of some interval from $F_n$ for some $n$. Please elaborate more!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 14:47












          • $begingroup$
            @LeAnhDung added some explanation.
            $endgroup$
            – Henno Brandsma
            Jan 19 at 14:56










          • $begingroup$
            Thank you so much! Your explanation is straight to my confusion and thus amazing!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 15:35














          1












          1








          1





          $begingroup$

          We remove the middle third of the interval to define the next $F_n$ and so it will be disjoint from $F_{n+1}$ so a fortiori from $F$.



          Explanation upon request: An interval from $F_n$ is of the form $[frac{k}{3^n}, frac{k+1}{3^n}]$ (not all such intervals are in $F_n$ but $2^n$ of them are; this makes the exact formula for $F-n$ tricky: specify which $k$ do occur; hence the recursive definition with sequences etc.) and we can also write this as $[frac{3k}{3^{n+1}}, frac{3(k+1)}{3^{n+1}}] = [frac{3k}{3^{n+1}}, frac{3k+3)}{3^{n+1}}]$, multiplying both parts of the fraction by $3$ and so its middle third open interval is $(frac{3k+1}{3^{n+1}}, frac{3k+2)}{3^{n+1}})$, exactly as claimed.






          share|cite|improve this answer











          $endgroup$



          We remove the middle third of the interval to define the next $F_n$ and so it will be disjoint from $F_{n+1}$ so a fortiori from $F$.



          Explanation upon request: An interval from $F_n$ is of the form $[frac{k}{3^n}, frac{k+1}{3^n}]$ (not all such intervals are in $F_n$ but $2^n$ of them are; this makes the exact formula for $F-n$ tricky: specify which $k$ do occur; hence the recursive definition with sequences etc.) and we can also write this as $[frac{3k}{3^{n+1}}, frac{3(k+1)}{3^{n+1}}] = [frac{3k}{3^{n+1}}, frac{3k+3)}{3^{n+1}}]$, multiplying both parts of the fraction by $3$ and so its middle third open interval is $(frac{3k+1}{3^{n+1}}, frac{3k+2)}{3^{n+1}})$, exactly as claimed.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 19 at 14:56

























          answered Jan 19 at 14:40









          Henno BrandsmaHenno Brandsma

          110k347116




          110k347116












          • $begingroup$
            From the definition of $F$, I am unable to infer that $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is the middle third of some interval from $F_n$ for some $n$. Please elaborate more!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 14:47












          • $begingroup$
            @LeAnhDung added some explanation.
            $endgroup$
            – Henno Brandsma
            Jan 19 at 14:56










          • $begingroup$
            Thank you so much! Your explanation is straight to my confusion and thus amazing!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 15:35


















          • $begingroup$
            From the definition of $F$, I am unable to infer that $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is the middle third of some interval from $F_n$ for some $n$. Please elaborate more!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 14:47












          • $begingroup$
            @LeAnhDung added some explanation.
            $endgroup$
            – Henno Brandsma
            Jan 19 at 14:56










          • $begingroup$
            Thank you so much! Your explanation is straight to my confusion and thus amazing!
            $endgroup$
            – Le Anh Dung
            Jan 19 at 15:35
















          $begingroup$
          From the definition of $F$, I am unable to infer that $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is the middle third of some interval from $F_n$ for some $n$. Please elaborate more!
          $endgroup$
          – Le Anh Dung
          Jan 19 at 14:47






          $begingroup$
          From the definition of $F$, I am unable to infer that $left(frac{3k+1}{3^{n+1}},,, frac{3k+2}{3^{n+1}}right)$ is the middle third of some interval from $F_n$ for some $n$. Please elaborate more!
          $endgroup$
          – Le Anh Dung
          Jan 19 at 14:47














          $begingroup$
          @LeAnhDung added some explanation.
          $endgroup$
          – Henno Brandsma
          Jan 19 at 14:56




          $begingroup$
          @LeAnhDung added some explanation.
          $endgroup$
          – Henno Brandsma
          Jan 19 at 14:56












          $begingroup$
          Thank you so much! Your explanation is straight to my confusion and thus amazing!
          $endgroup$
          – Le Anh Dung
          Jan 19 at 15:35




          $begingroup$
          Thank you so much! Your explanation is straight to my confusion and thus amazing!
          $endgroup$
          – Le Anh Dung
          Jan 19 at 15:35


















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