A “distinguishing” family of subsetsLower bound for the number of coin weightingsProve Intersection of Large Enough Subsets is Non-emptyMinimal collection of subsets to reconstruct singletonsUse of pigeonhole principle in ramsey-theorem about monochromatic triangles.Size of family $mathcal F = F_1, ldots, F_m$ is at least $lceil log_2nrceil$.Set theory question on subsets (considering pairs)How to use generalized pigeonhole principle to be sure that at least one of the integers picked is even?Proving that a set of 2016 natural numbers contain a non-empty set with a sum divisible by 2016The pigeonhole principle(?)Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$ board that are not on the same row.

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A “distinguishing” family of subsets


Lower bound for the number of coin weightingsProve Intersection of Large Enough Subsets is Non-emptyMinimal collection of subsets to reconstruct singletonsUse of pigeonhole principle in ramsey-theorem about monochromatic triangles.Size of family $mathcal F = F_1, ldots, F_m$ is at least $lceil log_2nrceil$.Set theory question on subsets (considering pairs)How to use generalized pigeonhole principle to be sure that at least one of the integers picked is even?Proving that a set of 2016 natural numbers contain a non-empty set with a sum divisible by 2016The pigeonhole principle(?)Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$ board that are not on the same row.













5












$begingroup$



Suppose $A$ is a finite set, $B$ is a collection of subsets of $A$, satisfying the following condition:



$$forall a, b in A, a neq b: exists C in B: (a in C) land (b notin C)$$



What is the least possible size of $B$.




Currently, I know that the minimal size of $B$ is not less than $lceil log_2 |A| rceil$ (by pigeonhole principle), and it does not exceed $2lceil log_2 |A| rceil$ (an example of that size can trivially be constructed). However, I do not know the exact answer to the question.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    But you surely mean $B subseteq mathfrak P(A)$, not $B in mathfrak P(A)$.
    $endgroup$
    – Andreas Caranti
    May 28 at 16:52















5












$begingroup$



Suppose $A$ is a finite set, $B$ is a collection of subsets of $A$, satisfying the following condition:



$$forall a, b in A, a neq b: exists C in B: (a in C) land (b notin C)$$



What is the least possible size of $B$.




Currently, I know that the minimal size of $B$ is not less than $lceil log_2 |A| rceil$ (by pigeonhole principle), and it does not exceed $2lceil log_2 |A| rceil$ (an example of that size can trivially be constructed). However, I do not know the exact answer to the question.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    But you surely mean $B subseteq mathfrak P(A)$, not $B in mathfrak P(A)$.
    $endgroup$
    – Andreas Caranti
    May 28 at 16:52













5












5








5


2



$begingroup$



Suppose $A$ is a finite set, $B$ is a collection of subsets of $A$, satisfying the following condition:



$$forall a, b in A, a neq b: exists C in B: (a in C) land (b notin C)$$



What is the least possible size of $B$.




Currently, I know that the minimal size of $B$ is not less than $lceil log_2 |A| rceil$ (by pigeonhole principle), and it does not exceed $2lceil log_2 |A| rceil$ (an example of that size can trivially be constructed). However, I do not know the exact answer to the question.










share|cite|improve this question











$endgroup$





Suppose $A$ is a finite set, $B$ is a collection of subsets of $A$, satisfying the following condition:



$$forall a, b in A, a neq b: exists C in B: (a in C) land (b notin C)$$



What is the least possible size of $B$.




Currently, I know that the minimal size of $B$ is not less than $lceil log_2 |A| rceil$ (by pigeonhole principle), and it does not exceed $2lceil log_2 |A| rceil$ (an example of that size can trivially be constructed). However, I do not know the exact answer to the question.







combinatorics discrete-mathematics elementary-set-theory pigeonhole-principle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 28 at 17:10







Yanior Weg

















asked May 28 at 16:32









Yanior WegYanior Weg

3,14531857




3,14531857







  • 1




    $begingroup$
    But you surely mean $B subseteq mathfrak P(A)$, not $B in mathfrak P(A)$.
    $endgroup$
    – Andreas Caranti
    May 28 at 16:52












  • 1




    $begingroup$
    But you surely mean $B subseteq mathfrak P(A)$, not $B in mathfrak P(A)$.
    $endgroup$
    – Andreas Caranti
    May 28 at 16:52







1




1




$begingroup$
But you surely mean $B subseteq mathfrak P(A)$, not $B in mathfrak P(A)$.
$endgroup$
– Andreas Caranti
May 28 at 16:52




$begingroup$
But you surely mean $B subseteq mathfrak P(A)$, not $B in mathfrak P(A)$.
$endgroup$
– Andreas Caranti
May 28 at 16:52










2 Answers
2






active

oldest

votes


















7












$begingroup$

The answer by @FabioSomenzi is not just an upperbound - it is actually tight.



  • Define $S(a) = C in B: a in C$ for all $a in A$. Note that $S(a) subset B$.


  • The main condition $forall aneq b: exists C in B: a in C land b notin C$ becomes $forall aneq b: S(a) notsubset S(b)$


  • Therefore the sets $S(a)_ain A$ form a Sperner family of size $|A|$.


  • The result now follows from Sperner's theorem, that the maximum sized Sperner family formed by subsets of $B$ has size $ choose $



UPDATE 2019-05-29: Lets see if we can get an estimate on the coefficient. Using Stirling's approximation $n! sim sqrt2 pi n (n over e)^n$, we have



$$n choose n/2 = n! over (n/2)! (n/2)! sim sqrt2 pi n (n over e)^n over sqrtpi n (n/2 over e)^n/2 sqrtpi n (n/2 over e)^n/2 = sqrt2 over pi n 2^n > |A|$$



So asymptotically we have $n sim log_2 |A|$, i.e. the coefficient is $1$, i.e. the pigeonhole-based lower bound is pretty tight.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    @YaniorWeg - much as I appreciate the checkmark :) the credit really goes to Fabio, both for coming up with the idea and for a great explanation. As soon as I understood his proof, it's immediately obvious that his row vectors form an antichain / Sperner family, and then it's just a matter of me having heard of Sperner's theorem.
    $endgroup$
    – antkam
    May 29 at 12:49


















6












$begingroup$

An improved upper bound is given by the smallest $n$ such that



$$ binomnlfloor n/2 rfloor geq |A| enspace. $$



For $A = a,b,c,d,e$, we get $n=4$ and, for example, the following encoding:
$$
beginarrayc
a & 1100 \
b & 1010 \
c & 1001 \
d & 0110 \
e & 0101
endarray
$$

Reading the columns of the table, $B_1 = a,b,c, B_2 = a,d,e, B_3 = b,d, B_4 = c,e$. Since each element of $A$ appears in the same number of subsets, and no two elements appear in the same subsets, the condition is satisfied.






share|cite|improve this answer









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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    The answer by @FabioSomenzi is not just an upperbound - it is actually tight.



    • Define $S(a) = C in B: a in C$ for all $a in A$. Note that $S(a) subset B$.


    • The main condition $forall aneq b: exists C in B: a in C land b notin C$ becomes $forall aneq b: S(a) notsubset S(b)$


    • Therefore the sets $S(a)_ain A$ form a Sperner family of size $|A|$.


    • The result now follows from Sperner's theorem, that the maximum sized Sperner family formed by subsets of $B$ has size $ choose $



    UPDATE 2019-05-29: Lets see if we can get an estimate on the coefficient. Using Stirling's approximation $n! sim sqrt2 pi n (n over e)^n$, we have



    $$n choose n/2 = n! over (n/2)! (n/2)! sim sqrt2 pi n (n over e)^n over sqrtpi n (n/2 over e)^n/2 sqrtpi n (n/2 over e)^n/2 = sqrt2 over pi n 2^n > |A|$$



    So asymptotically we have $n sim log_2 |A|$, i.e. the coefficient is $1$, i.e. the pigeonhole-based lower bound is pretty tight.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      @YaniorWeg - much as I appreciate the checkmark :) the credit really goes to Fabio, both for coming up with the idea and for a great explanation. As soon as I understood his proof, it's immediately obvious that his row vectors form an antichain / Sperner family, and then it's just a matter of me having heard of Sperner's theorem.
      $endgroup$
      – antkam
      May 29 at 12:49















    7












    $begingroup$

    The answer by @FabioSomenzi is not just an upperbound - it is actually tight.



    • Define $S(a) = C in B: a in C$ for all $a in A$. Note that $S(a) subset B$.


    • The main condition $forall aneq b: exists C in B: a in C land b notin C$ becomes $forall aneq b: S(a) notsubset S(b)$


    • Therefore the sets $S(a)_ain A$ form a Sperner family of size $|A|$.


    • The result now follows from Sperner's theorem, that the maximum sized Sperner family formed by subsets of $B$ has size $ choose $



    UPDATE 2019-05-29: Lets see if we can get an estimate on the coefficient. Using Stirling's approximation $n! sim sqrt2 pi n (n over e)^n$, we have



    $$n choose n/2 = n! over (n/2)! (n/2)! sim sqrt2 pi n (n over e)^n over sqrtpi n (n/2 over e)^n/2 sqrtpi n (n/2 over e)^n/2 = sqrt2 over pi n 2^n > |A|$$



    So asymptotically we have $n sim log_2 |A|$, i.e. the coefficient is $1$, i.e. the pigeonhole-based lower bound is pretty tight.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      @YaniorWeg - much as I appreciate the checkmark :) the credit really goes to Fabio, both for coming up with the idea and for a great explanation. As soon as I understood his proof, it's immediately obvious that his row vectors form an antichain / Sperner family, and then it's just a matter of me having heard of Sperner's theorem.
      $endgroup$
      – antkam
      May 29 at 12:49













    7












    7








    7





    $begingroup$

    The answer by @FabioSomenzi is not just an upperbound - it is actually tight.



    • Define $S(a) = C in B: a in C$ for all $a in A$. Note that $S(a) subset B$.


    • The main condition $forall aneq b: exists C in B: a in C land b notin C$ becomes $forall aneq b: S(a) notsubset S(b)$


    • Therefore the sets $S(a)_ain A$ form a Sperner family of size $|A|$.


    • The result now follows from Sperner's theorem, that the maximum sized Sperner family formed by subsets of $B$ has size $ choose $



    UPDATE 2019-05-29: Lets see if we can get an estimate on the coefficient. Using Stirling's approximation $n! sim sqrt2 pi n (n over e)^n$, we have



    $$n choose n/2 = n! over (n/2)! (n/2)! sim sqrt2 pi n (n over e)^n over sqrtpi n (n/2 over e)^n/2 sqrtpi n (n/2 over e)^n/2 = sqrt2 over pi n 2^n > |A|$$



    So asymptotically we have $n sim log_2 |A|$, i.e. the coefficient is $1$, i.e. the pigeonhole-based lower bound is pretty tight.






    share|cite|improve this answer











    $endgroup$



    The answer by @FabioSomenzi is not just an upperbound - it is actually tight.



    • Define $S(a) = C in B: a in C$ for all $a in A$. Note that $S(a) subset B$.


    • The main condition $forall aneq b: exists C in B: a in C land b notin C$ becomes $forall aneq b: S(a) notsubset S(b)$


    • Therefore the sets $S(a)_ain A$ form a Sperner family of size $|A|$.


    • The result now follows from Sperner's theorem, that the maximum sized Sperner family formed by subsets of $B$ has size $ choose $



    UPDATE 2019-05-29: Lets see if we can get an estimate on the coefficient. Using Stirling's approximation $n! sim sqrt2 pi n (n over e)^n$, we have



    $$n choose n/2 = n! over (n/2)! (n/2)! sim sqrt2 pi n (n over e)^n over sqrtpi n (n/2 over e)^n/2 sqrtpi n (n/2 over e)^n/2 = sqrt2 over pi n 2^n > |A|$$



    So asymptotically we have $n sim log_2 |A|$, i.e. the coefficient is $1$, i.e. the pigeonhole-based lower bound is pretty tight.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 29 at 13:07

























    answered May 28 at 23:39









    antkamantkam

    5,174415




    5,174415











    • $begingroup$
      @YaniorWeg - much as I appreciate the checkmark :) the credit really goes to Fabio, both for coming up with the idea and for a great explanation. As soon as I understood his proof, it's immediately obvious that his row vectors form an antichain / Sperner family, and then it's just a matter of me having heard of Sperner's theorem.
      $endgroup$
      – antkam
      May 29 at 12:49
















    • $begingroup$
      @YaniorWeg - much as I appreciate the checkmark :) the credit really goes to Fabio, both for coming up with the idea and for a great explanation. As soon as I understood his proof, it's immediately obvious that his row vectors form an antichain / Sperner family, and then it's just a matter of me having heard of Sperner's theorem.
      $endgroup$
      – antkam
      May 29 at 12:49















    $begingroup$
    @YaniorWeg - much as I appreciate the checkmark :) the credit really goes to Fabio, both for coming up with the idea and for a great explanation. As soon as I understood his proof, it's immediately obvious that his row vectors form an antichain / Sperner family, and then it's just a matter of me having heard of Sperner's theorem.
    $endgroup$
    – antkam
    May 29 at 12:49




    $begingroup$
    @YaniorWeg - much as I appreciate the checkmark :) the credit really goes to Fabio, both for coming up with the idea and for a great explanation. As soon as I understood his proof, it's immediately obvious that his row vectors form an antichain / Sperner family, and then it's just a matter of me having heard of Sperner's theorem.
    $endgroup$
    – antkam
    May 29 at 12:49











    6












    $begingroup$

    An improved upper bound is given by the smallest $n$ such that



    $$ binomnlfloor n/2 rfloor geq |A| enspace. $$



    For $A = a,b,c,d,e$, we get $n=4$ and, for example, the following encoding:
    $$
    beginarrayc
    a & 1100 \
    b & 1010 \
    c & 1001 \
    d & 0110 \
    e & 0101
    endarray
    $$

    Reading the columns of the table, $B_1 = a,b,c, B_2 = a,d,e, B_3 = b,d, B_4 = c,e$. Since each element of $A$ appears in the same number of subsets, and no two elements appear in the same subsets, the condition is satisfied.






    share|cite|improve this answer









    $endgroup$

















      6












      $begingroup$

      An improved upper bound is given by the smallest $n$ such that



      $$ binomnlfloor n/2 rfloor geq |A| enspace. $$



      For $A = a,b,c,d,e$, we get $n=4$ and, for example, the following encoding:
      $$
      beginarrayc
      a & 1100 \
      b & 1010 \
      c & 1001 \
      d & 0110 \
      e & 0101
      endarray
      $$

      Reading the columns of the table, $B_1 = a,b,c, B_2 = a,d,e, B_3 = b,d, B_4 = c,e$. Since each element of $A$ appears in the same number of subsets, and no two elements appear in the same subsets, the condition is satisfied.






      share|cite|improve this answer









      $endgroup$















        6












        6








        6





        $begingroup$

        An improved upper bound is given by the smallest $n$ such that



        $$ binomnlfloor n/2 rfloor geq |A| enspace. $$



        For $A = a,b,c,d,e$, we get $n=4$ and, for example, the following encoding:
        $$
        beginarrayc
        a & 1100 \
        b & 1010 \
        c & 1001 \
        d & 0110 \
        e & 0101
        endarray
        $$

        Reading the columns of the table, $B_1 = a,b,c, B_2 = a,d,e, B_3 = b,d, B_4 = c,e$. Since each element of $A$ appears in the same number of subsets, and no two elements appear in the same subsets, the condition is satisfied.






        share|cite|improve this answer









        $endgroup$



        An improved upper bound is given by the smallest $n$ such that



        $$ binomnlfloor n/2 rfloor geq |A| enspace. $$



        For $A = a,b,c,d,e$, we get $n=4$ and, for example, the following encoding:
        $$
        beginarrayc
        a & 1100 \
        b & 1010 \
        c & 1001 \
        d & 0110 \
        e & 0101
        endarray
        $$

        Reading the columns of the table, $B_1 = a,b,c, B_2 = a,d,e, B_3 = b,d, B_4 = c,e$. Since each element of $A$ appears in the same number of subsets, and no two elements appear in the same subsets, the condition is satisfied.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 28 at 17:56









        Fabio SomenziFabio Somenzi

        6,81121321




        6,81121321



























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