Rudin's proof that the Cantor set has no segments












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In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.






real-analysis cantor-set







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    3












    $begingroup$


    In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.






    real-analysis cantor-set







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      3



      $begingroup$


      In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.






      real-analysis cantor-set







      share|cite|improve this question











      $endgroup$




      In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.






      real-analysis cantor-set




      real-analysis cantor-set






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 12 '15 at 18:14







      Milad

















      asked Oct 10 '15 at 19:03









      MiladMilad

      819




      819






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Your denominators should be $3^n$, not $3^{-n}$.



          Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is



          $$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$



          if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then



          $$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$



          so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that



          $$frac1{3^n}lefrac{beta-alpha}4;,$$



          the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.



          Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @Milad: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:24










          • $begingroup$
            in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
            $endgroup$
            – Milad
            Oct 10 '15 at 19:42






          • 1




            $begingroup$
            @Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:45










          • $begingroup$
            @Milad: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:59



















          5












          $begingroup$

          We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,



          so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.



          Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so



          $(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Very nice proof!
            $endgroup$
            – ZFR
            Oct 10 '15 at 19:22










          • $begingroup$
            @RFZ Thanks for your comment; I appreciate it!
            $endgroup$
            – user84413
            Oct 10 '15 at 19:24










          • $begingroup$
            You are always welcome!
            $endgroup$
            – ZFR
            Oct 10 '15 at 19:25










          • $begingroup$
            your proof is really smart. thanks @user84413
            $endgroup$
            – Milad
            Oct 10 '15 at 19:35



















          0












          $begingroup$

          enter image description here



          If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.

          See the above picture.






          share|cite|improve this answer









          $endgroup$














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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Your denominators should be $3^n$, not $3^{-n}$.



            Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is



            $$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$



            if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then



            $$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$



            so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that



            $$frac1{3^n}lefrac{beta-alpha}4;,$$



            the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.



            Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:24










            • $begingroup$
              in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
              $endgroup$
              – Milad
              Oct 10 '15 at 19:42






            • 1




              $begingroup$
              @Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:45










            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:59
















            4












            $begingroup$

            Your denominators should be $3^n$, not $3^{-n}$.



            Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is



            $$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$



            if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then



            $$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$



            so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that



            $$frac1{3^n}lefrac{beta-alpha}4;,$$



            the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.



            Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:24










            • $begingroup$
              in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
              $endgroup$
              – Milad
              Oct 10 '15 at 19:42






            • 1




              $begingroup$
              @Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:45










            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:59














            4












            4








            4





            $begingroup$

            Your denominators should be $3^n$, not $3^{-n}$.



            Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is



            $$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$



            if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then



            $$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$



            so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that



            $$frac1{3^n}lefrac{beta-alpha}4;,$$



            the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.



            Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.






            share|cite|improve this answer









            $endgroup$



            Your denominators should be $3^n$, not $3^{-n}$.



            Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is



            $$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$



            if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then



            $$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$



            so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that



            $$frac1{3^n}lefrac{beta-alpha}4;,$$



            the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.



            Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 10 '15 at 19:17









            Brian M. ScottBrian M. Scott

            460k40517918




            460k40517918












            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:24










            • $begingroup$
              in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
              $endgroup$
              – Milad
              Oct 10 '15 at 19:42






            • 1




              $begingroup$
              @Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:45










            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:59


















            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:24










            • $begingroup$
              in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
              $endgroup$
              – Milad
              Oct 10 '15 at 19:42






            • 1




              $begingroup$
              @Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:45










            • $begingroup$
              @Milad: You’re welcome.
              $endgroup$
              – Brian M. Scott
              Oct 10 '15 at 19:59
















            $begingroup$
            @Milad: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:24




            $begingroup$
            @Milad: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:24












            $begingroup$
            in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
            $endgroup$
            – Milad
            Oct 10 '15 at 19:42




            $begingroup$
            in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
            $endgroup$
            – Milad
            Oct 10 '15 at 19:42




            1




            1




            $begingroup$
            @Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:45




            $begingroup$
            @Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:45












            $begingroup$
            @Milad: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:59




            $begingroup$
            @Milad: You’re welcome.
            $endgroup$
            – Brian M. Scott
            Oct 10 '15 at 19:59











            5












            $begingroup$

            We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,



            so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.



            Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so



            $(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Very nice proof!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:22










            • $begingroup$
              @RFZ Thanks for your comment; I appreciate it!
              $endgroup$
              – user84413
              Oct 10 '15 at 19:24










            • $begingroup$
              You are always welcome!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:25










            • $begingroup$
              your proof is really smart. thanks @user84413
              $endgroup$
              – Milad
              Oct 10 '15 at 19:35
















            5












            $begingroup$

            We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,



            so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.



            Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so



            $(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Very nice proof!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:22










            • $begingroup$
              @RFZ Thanks for your comment; I appreciate it!
              $endgroup$
              – user84413
              Oct 10 '15 at 19:24










            • $begingroup$
              You are always welcome!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:25










            • $begingroup$
              your proof is really smart. thanks @user84413
              $endgroup$
              – Milad
              Oct 10 '15 at 19:35














            5












            5








            5





            $begingroup$

            We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,



            so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.



            Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so



            $(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.






            share|cite|improve this answer











            $endgroup$



            We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,



            so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.



            Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so



            $(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 10 '15 at 19:22

























            answered Oct 10 '15 at 19:16









            user84413user84413

            23k11848




            23k11848












            • $begingroup$
              Very nice proof!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:22










            • $begingroup$
              @RFZ Thanks for your comment; I appreciate it!
              $endgroup$
              – user84413
              Oct 10 '15 at 19:24










            • $begingroup$
              You are always welcome!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:25










            • $begingroup$
              your proof is really smart. thanks @user84413
              $endgroup$
              – Milad
              Oct 10 '15 at 19:35


















            • $begingroup$
              Very nice proof!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:22










            • $begingroup$
              @RFZ Thanks for your comment; I appreciate it!
              $endgroup$
              – user84413
              Oct 10 '15 at 19:24










            • $begingroup$
              You are always welcome!
              $endgroup$
              – ZFR
              Oct 10 '15 at 19:25










            • $begingroup$
              your proof is really smart. thanks @user84413
              $endgroup$
              – Milad
              Oct 10 '15 at 19:35
















            $begingroup$
            Very nice proof!
            $endgroup$
            – ZFR
            Oct 10 '15 at 19:22




            $begingroup$
            Very nice proof!
            $endgroup$
            – ZFR
            Oct 10 '15 at 19:22












            $begingroup$
            @RFZ Thanks for your comment; I appreciate it!
            $endgroup$
            – user84413
            Oct 10 '15 at 19:24




            $begingroup$
            @RFZ Thanks for your comment; I appreciate it!
            $endgroup$
            – user84413
            Oct 10 '15 at 19:24












            $begingroup$
            You are always welcome!
            $endgroup$
            – ZFR
            Oct 10 '15 at 19:25




            $begingroup$
            You are always welcome!
            $endgroup$
            – ZFR
            Oct 10 '15 at 19:25












            $begingroup$
            your proof is really smart. thanks @user84413
            $endgroup$
            – Milad
            Oct 10 '15 at 19:35




            $begingroup$
            your proof is really smart. thanks @user84413
            $endgroup$
            – Milad
            Oct 10 '15 at 19:35











            0












            $begingroup$

            enter image description here



            If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.

            See the above picture.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              enter image description here



              If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.

              See the above picture.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                enter image description here



                If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.

                See the above picture.






                share|cite|improve this answer









                $endgroup$



                enter image description here



                If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.

                See the above picture.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 1:31









                tchappy hatchappy ha

                778412




                778412






























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