How to prove second part of this combinatorial lemma?

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Lemma is stated like this:



Lemma 1. Suppose that a finite number of points subdivides a closed interval
into smaller intervals. The left endpoint of the original interval is labeled by
0, the right one by 1, and each of the partitioning points inside the interval
is also labeled by either 0 or 1. Then there is an interval of the subdivision
whose endpoints are labeled by different numbers. Moreover, the number of
such intervals is odd.



I understand existence part, that is, that there is an interval of the subdivision
whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?



How would you prove that?







share|cite|improve this question
























    up vote
    1
    down vote

    favorite












    Lemma is stated like this:



    Lemma 1. Suppose that a finite number of points subdivides a closed interval
    into smaller intervals. The left endpoint of the original interval is labeled by
    0, the right one by 1, and each of the partitioning points inside the interval
    is also labeled by either 0 or 1. Then there is an interval of the subdivision
    whose endpoints are labeled by different numbers. Moreover, the number of
    such intervals is odd.



    I understand existence part, that is, that there is an interval of the subdivision
    whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?



    How would you prove that?







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Lemma is stated like this:



      Lemma 1. Suppose that a finite number of points subdivides a closed interval
      into smaller intervals. The left endpoint of the original interval is labeled by
      0, the right one by 1, and each of the partitioning points inside the interval
      is also labeled by either 0 or 1. Then there is an interval of the subdivision
      whose endpoints are labeled by different numbers. Moreover, the number of
      such intervals is odd.



      I understand existence part, that is, that there is an interval of the subdivision
      whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?



      How would you prove that?







      share|cite|improve this question












      Lemma is stated like this:



      Lemma 1. Suppose that a finite number of points subdivides a closed interval
      into smaller intervals. The left endpoint of the original interval is labeled by
      0, the right one by 1, and each of the partitioning points inside the interval
      is also labeled by either 0 or 1. Then there is an interval of the subdivision
      whose endpoints are labeled by different numbers. Moreover, the number of
      such intervals is odd.



      I understand existence part, that is, that there is an interval of the subdivision
      whose endpoints are labeled by different numbers. But I do not understand why the number of such intervals must be odd?



      How would you prove that?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 24 at 9:32









      Right

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      1,026213




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote













          If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).



          But that number is precisely the number of intervals with different labels.



          By the way this is the toy version of Sperner's lemma.






          share|cite|improve this answer


















          • 1




            Perhaps insert or vice versa in the first clause for clarity?
            – Peter Taylor
            Aug 24 at 10:48










          • @PeterTaylor Good point. Thanks!
            – Michal Adamaszek
            Aug 24 at 19:34

















          up vote
          0
          down vote













          You can use induction to prove that.



          Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.



          This enforces your induction hypothese and makes things more easy to prove.



          The statement is obviously true if $n=0$ points are added.



          If $n>0$ points are added then pick out one of these (inner) points.



          It splits up the original interval in two intervals with $<n$ points.



          Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.



          That leads to the expected conclusions for $n$.






          share|cite|improve this answer






















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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            up vote
            1
            down vote













            If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).



            But that number is precisely the number of intervals with different labels.



            By the way this is the toy version of Sperner's lemma.






            share|cite|improve this answer


















            • 1




              Perhaps insert or vice versa in the first clause for clarity?
              – Peter Taylor
              Aug 24 at 10:48










            • @PeterTaylor Good point. Thanks!
              – Michal Adamaszek
              Aug 24 at 19:34














            up vote
            1
            down vote













            If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).



            But that number is precisely the number of intervals with different labels.



            By the way this is the toy version of Sperner's lemma.






            share|cite|improve this answer


















            • 1




              Perhaps insert or vice versa in the first clause for clarity?
              – Peter Taylor
              Aug 24 at 10:48










            • @PeterTaylor Good point. Thanks!
              – Michal Adamaszek
              Aug 24 at 19:34












            up vote
            1
            down vote










            up vote
            1
            down vote









            If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).



            But that number is precisely the number of intervals with different labels.



            By the way this is the toy version of Sperner's lemma.






            share|cite|improve this answer














            If you go from left to right and record the number of times 0 changed to 1 or vice versa, then this number is odd (because you start with 0 and end with 1).



            But that number is precisely the number of intervals with different labels.



            By the way this is the toy version of Sperner's lemma.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 24 at 19:33

























            answered Aug 24 at 9:54









            Michal Adamaszek

            1,69948




            1,69948







            • 1




              Perhaps insert or vice versa in the first clause for clarity?
              – Peter Taylor
              Aug 24 at 10:48










            • @PeterTaylor Good point. Thanks!
              – Michal Adamaszek
              Aug 24 at 19:34












            • 1




              Perhaps insert or vice versa in the first clause for clarity?
              – Peter Taylor
              Aug 24 at 10:48










            • @PeterTaylor Good point. Thanks!
              – Michal Adamaszek
              Aug 24 at 19:34







            1




            1




            Perhaps insert or vice versa in the first clause for clarity?
            – Peter Taylor
            Aug 24 at 10:48




            Perhaps insert or vice versa in the first clause for clarity?
            – Peter Taylor
            Aug 24 at 10:48












            @PeterTaylor Good point. Thanks!
            – Michal Adamaszek
            Aug 24 at 19:34




            @PeterTaylor Good point. Thanks!
            – Michal Adamaszek
            Aug 24 at 19:34










            up vote
            0
            down vote













            You can use induction to prove that.



            Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.



            This enforces your induction hypothese and makes things more easy to prove.



            The statement is obviously true if $n=0$ points are added.



            If $n>0$ points are added then pick out one of these (inner) points.



            It splits up the original interval in two intervals with $<n$ points.



            Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.



            That leads to the expected conclusions for $n$.






            share|cite|improve this answer


























              up vote
              0
              down vote













              You can use induction to prove that.



              Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.



              This enforces your induction hypothese and makes things more easy to prove.



              The statement is obviously true if $n=0$ points are added.



              If $n>0$ points are added then pick out one of these (inner) points.



              It splits up the original interval in two intervals with $<n$ points.



              Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.



              That leads to the expected conclusions for $n$.






              share|cite|improve this answer
























                up vote
                0
                down vote










                up vote
                0
                down vote









                You can use induction to prove that.



                Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.



                This enforces your induction hypothese and makes things more easy to prove.



                The statement is obviously true if $n=0$ points are added.



                If $n>0$ points are added then pick out one of these (inner) points.



                It splits up the original interval in two intervals with $<n$ points.



                Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.



                That leads to the expected conclusions for $n$.






                share|cite|improve this answer














                You can use induction to prove that.



                Extend the statement by adding that the number of such intervals is even if the points of the original interval are both labeled by $0$ or are both labeled by $1$.



                This enforces your induction hypothese and makes things more easy to prove.



                The statement is obviously true if $n=0$ points are added.



                If $n>0$ points are added then pick out one of these (inner) points.



                It splits up the original interval in two intervals with $<n$ points.



                Treating them separatedly for both of them the induction hypothese informs you completely about the question whether the number of the special intervals are odd or even.



                That leads to the expected conclusions for $n$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Aug 24 at 13:06

























                answered Aug 24 at 10:01









                drhab

                88.3k541120




                88.3k541120



























                     

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