Uniqueness of set where subset exactly sums to 1












1














Let $A = {a_1, a_2, ldots, a_I}$ be a set of real numbers, where for all $i in {1,2,ldots,I}$, $0<a_ile1$ and $sum_{j ne i} a_j ge 1$.



I am interested in the set (or sets) $A$ satisfying the following condition:




For every subset $S subset {1,2,ldots,I}$ where $sum_{i in S} a_i ge 1$, there exists a subset $S' subseteq S$ where $sum_{i in S'} a_i = 1$, i.e. that exactly sums to 1.




Is it true that the only set $A$ satisfying this condition is the one where, for all $i in {1,2,ldots,I}$:
$$a_i = a quadlandquad frac{1}{a} text{ is an integer}$$






proof-verification integers







share|cite|improve this question






















  • Could you please give an example of what you're looking for?
    – MJD
    Nov 23 '18 at 14:03










  • @MJD: $A=left{frac{1}{2}, frac{1}{2}, frac{1}{2}right}$ satisfies the condition, while $A=left{frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ doesn't since the sum of $S'=left{frac{1}{3}, frac{1}{2}right}$ does not exactly equal 1.
    – bonna
    Nov 23 '18 at 14:14










  • But for that second example, $A$ doesn't satisfy the $forall iin {1,ldots,I}: sum_{jneq i} a_j geq 1$?
    – Henrik
    Nov 23 '18 at 14:16












  • Sorry. Let the second example be $A=left{frac{1}{3}, frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ and $S'=left{frac{1}{3}, frac{1}{3}, frac{1}{2}right}$
    – bonna
    Nov 23 '18 at 14:20


















1














Let $A = {a_1, a_2, ldots, a_I}$ be a set of real numbers, where for all $i in {1,2,ldots,I}$, $0<a_ile1$ and $sum_{j ne i} a_j ge 1$.



I am interested in the set (or sets) $A$ satisfying the following condition:




For every subset $S subset {1,2,ldots,I}$ where $sum_{i in S} a_i ge 1$, there exists a subset $S' subseteq S$ where $sum_{i in S'} a_i = 1$, i.e. that exactly sums to 1.




Is it true that the only set $A$ satisfying this condition is the one where, for all $i in {1,2,ldots,I}$:
$$a_i = a quadlandquad frac{1}{a} text{ is an integer}$$






proof-verification integers







share|cite|improve this question






















  • Could you please give an example of what you're looking for?
    – MJD
    Nov 23 '18 at 14:03










  • @MJD: $A=left{frac{1}{2}, frac{1}{2}, frac{1}{2}right}$ satisfies the condition, while $A=left{frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ doesn't since the sum of $S'=left{frac{1}{3}, frac{1}{2}right}$ does not exactly equal 1.
    – bonna
    Nov 23 '18 at 14:14










  • But for that second example, $A$ doesn't satisfy the $forall iin {1,ldots,I}: sum_{jneq i} a_j geq 1$?
    – Henrik
    Nov 23 '18 at 14:16












  • Sorry. Let the second example be $A=left{frac{1}{3}, frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ and $S'=left{frac{1}{3}, frac{1}{3}, frac{1}{2}right}$
    – bonna
    Nov 23 '18 at 14:20
















1












1








1







Let $A = {a_1, a_2, ldots, a_I}$ be a set of real numbers, where for all $i in {1,2,ldots,I}$, $0<a_ile1$ and $sum_{j ne i} a_j ge 1$.



I am interested in the set (or sets) $A$ satisfying the following condition:




For every subset $S subset {1,2,ldots,I}$ where $sum_{i in S} a_i ge 1$, there exists a subset $S' subseteq S$ where $sum_{i in S'} a_i = 1$, i.e. that exactly sums to 1.




Is it true that the only set $A$ satisfying this condition is the one where, for all $i in {1,2,ldots,I}$:
$$a_i = a quadlandquad frac{1}{a} text{ is an integer}$$






proof-verification integers







share|cite|improve this question













Let $A = {a_1, a_2, ldots, a_I}$ be a set of real numbers, where for all $i in {1,2,ldots,I}$, $0<a_ile1$ and $sum_{j ne i} a_j ge 1$.



I am interested in the set (or sets) $A$ satisfying the following condition:




For every subset $S subset {1,2,ldots,I}$ where $sum_{i in S} a_i ge 1$, there exists a subset $S' subseteq S$ where $sum_{i in S'} a_i = 1$, i.e. that exactly sums to 1.




Is it true that the only set $A$ satisfying this condition is the one where, for all $i in {1,2,ldots,I}$:
$$a_i = a quadlandquad frac{1}{a} text{ is an integer}$$






proof-verification integers




proof-verification integers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 23 '18 at 13:58









bonnabonna

858




858












  • Could you please give an example of what you're looking for?
    – MJD
    Nov 23 '18 at 14:03










  • @MJD: $A=left{frac{1}{2}, frac{1}{2}, frac{1}{2}right}$ satisfies the condition, while $A=left{frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ doesn't since the sum of $S'=left{frac{1}{3}, frac{1}{2}right}$ does not exactly equal 1.
    – bonna
    Nov 23 '18 at 14:14










  • But for that second example, $A$ doesn't satisfy the $forall iin {1,ldots,I}: sum_{jneq i} a_j geq 1$?
    – Henrik
    Nov 23 '18 at 14:16












  • Sorry. Let the second example be $A=left{frac{1}{3}, frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ and $S'=left{frac{1}{3}, frac{1}{3}, frac{1}{2}right}$
    – bonna
    Nov 23 '18 at 14:20




















  • Could you please give an example of what you're looking for?
    – MJD
    Nov 23 '18 at 14:03










  • @MJD: $A=left{frac{1}{2}, frac{1}{2}, frac{1}{2}right}$ satisfies the condition, while $A=left{frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ doesn't since the sum of $S'=left{frac{1}{3}, frac{1}{2}right}$ does not exactly equal 1.
    – bonna
    Nov 23 '18 at 14:14










  • But for that second example, $A$ doesn't satisfy the $forall iin {1,ldots,I}: sum_{jneq i} a_j geq 1$?
    – Henrik
    Nov 23 '18 at 14:16












  • Sorry. Let the second example be $A=left{frac{1}{3}, frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ and $S'=left{frac{1}{3}, frac{1}{3}, frac{1}{2}right}$
    – bonna
    Nov 23 '18 at 14:20


















Could you please give an example of what you're looking for?
– MJD
Nov 23 '18 at 14:03




Could you please give an example of what you're looking for?
– MJD
Nov 23 '18 at 14:03












@MJD: $A=left{frac{1}{2}, frac{1}{2}, frac{1}{2}right}$ satisfies the condition, while $A=left{frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ doesn't since the sum of $S'=left{frac{1}{3}, frac{1}{2}right}$ does not exactly equal 1.
– bonna
Nov 23 '18 at 14:14




@MJD: $A=left{frac{1}{2}, frac{1}{2}, frac{1}{2}right}$ satisfies the condition, while $A=left{frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ doesn't since the sum of $S'=left{frac{1}{3}, frac{1}{2}right}$ does not exactly equal 1.
– bonna
Nov 23 '18 at 14:14












But for that second example, $A$ doesn't satisfy the $forall iin {1,ldots,I}: sum_{jneq i} a_j geq 1$?
– Henrik
Nov 23 '18 at 14:16






But for that second example, $A$ doesn't satisfy the $forall iin {1,ldots,I}: sum_{jneq i} a_j geq 1$?
– Henrik
Nov 23 '18 at 14:16














Sorry. Let the second example be $A=left{frac{1}{3}, frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ and $S'=left{frac{1}{3}, frac{1}{3}, frac{1}{2}right}$
– bonna
Nov 23 '18 at 14:20






Sorry. Let the second example be $A=left{frac{1}{3}, frac{1}{3}, frac{1}{2}, frac{1}{2}right}$ and $S'=left{frac{1}{3}, frac{1}{3}, frac{1}{2}right}$
– bonna
Nov 23 '18 at 14:20












1 Answer
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No, it is not true. Consider the following counter-example where $a_i ne a, forall i$:



$$A = {a_1, a_2, a_3, a_4}, text{ where } a_1 = a_2 = frac{1}{4} text{ and } a_3 = a_4 = frac{1}{2}$$



This satisfies the condition that all the subsets $$S = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}, {a_1, a_3, a_4}, {a_2, a_3, a_4}$$ where $sum_{i in S} a_i ge 1$, have subsets $$S' = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}$$ where $sum_{i in S'} a_i = 1$.






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    1 Answer
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    active

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    No, it is not true. Consider the following counter-example where $a_i ne a, forall i$:



    $$A = {a_1, a_2, a_3, a_4}, text{ where } a_1 = a_2 = frac{1}{4} text{ and } a_3 = a_4 = frac{1}{2}$$



    This satisfies the condition that all the subsets $$S = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}, {a_1, a_3, a_4}, {a_2, a_3, a_4}$$ where $sum_{i in S} a_i ge 1$, have subsets $$S' = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}$$ where $sum_{i in S'} a_i = 1$.






    share|cite|improve this answer


























      0














      No, it is not true. Consider the following counter-example where $a_i ne a, forall i$:



      $$A = {a_1, a_2, a_3, a_4}, text{ where } a_1 = a_2 = frac{1}{4} text{ and } a_3 = a_4 = frac{1}{2}$$



      This satisfies the condition that all the subsets $$S = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}, {a_1, a_3, a_4}, {a_2, a_3, a_4}$$ where $sum_{i in S} a_i ge 1$, have subsets $$S' = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}$$ where $sum_{i in S'} a_i = 1$.






      share|cite|improve this answer
























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        0






        No, it is not true. Consider the following counter-example where $a_i ne a, forall i$:



        $$A = {a_1, a_2, a_3, a_4}, text{ where } a_1 = a_2 = frac{1}{4} text{ and } a_3 = a_4 = frac{1}{2}$$



        This satisfies the condition that all the subsets $$S = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}, {a_1, a_3, a_4}, {a_2, a_3, a_4}$$ where $sum_{i in S} a_i ge 1$, have subsets $$S' = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}$$ where $sum_{i in S'} a_i = 1$.






        share|cite|improve this answer












        No, it is not true. Consider the following counter-example where $a_i ne a, forall i$:



        $$A = {a_1, a_2, a_3, a_4}, text{ where } a_1 = a_2 = frac{1}{4} text{ and } a_3 = a_4 = frac{1}{2}$$



        This satisfies the condition that all the subsets $$S = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}, {a_1, a_3, a_4}, {a_2, a_3, a_4}$$ where $sum_{i in S} a_i ge 1$, have subsets $$S' = {a_1, a_2, a_3}, {a_1, a_2, a_4}, {a_3, a_4}$$ where $sum_{i in S'} a_i = 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 '18 at 15:12









        bonnabonna

        858




        858






























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