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Consider two sets $A,B$ composed of two real numbers each.

These four real numbers are in $[0,1]$.

Consider other two real numbers $c\in [0,1]$, $d\in [0,1]$.

Assume there exists a way of ordering the two numbers in each set $A,B$ such that $$ \begin{cases} w^A_1+w^B_1=c\\ w^A_2+w^B_2=d \end{cases} $$ where

  • $w^A_h$ denotes the $h$th element of set $A$ once we have ordered its two elements

  • $w^B_h$ denotes the $h$th element of set $B$ once we have ordered its two element

Claim: if such an ordering is not unique, then it should be that the two numbers in $A$ are equal and/or that the two numbers in $B$ are equal.

Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?

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    $\begingroup$ If such an ordering exists, we have $w_2^A+w_2^B=d=1-c=1-w_1^A-w_1^B\implies w_1^B+w_2^B=0\implies B=\{0,0\}$ which is a contradiction. So the ordering doesn't exist. $\endgroup$ Dec 22, 2018 at 19:14
  • $\begingroup$ Yes, sorry, I have deleted the summing up to one. Thanks for the observation. $\endgroup$
    – TEX
    Dec 22, 2018 at 19:30

1 Answer 1

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Say the ordering is not unique. Then at-least one of $A,B$ can be ordered in two ways. Without loss of generality, let us assume that set is $A$. Keeping the order of $B$ intact, we have

$\implies w_1^A+w_1^B=w_2^A+w_1^B=c\implies w_1^A=w_2^A$

Therefore, both the elements of $A$ are identical. We can assume that the ordering of $B$ is not unique and land at a similar conclusion for $B$.

Edit. As suggested by the OP, I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings changed. In that case, $c=d$ and the elements of $A,B$ need not be equal.

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  • $\begingroup$ It could be that $w_1^A+w_1^B=c$ and $w_2^A+w_2^B=d$ and that $w_1^A+w_1^B=d$ and $w_2^A+w_2^B=c$ which implies $c=d$ (but the elements of $A$ can be different between each other and the elements of $B$ can be different between each other). $\endgroup$
    – TEX
    Dec 22, 2018 at 20:02
  • $\begingroup$ Is it fair to say that if $c\neq d$ then your proof is correct? $\endgroup$
    – TEX
    Dec 22, 2018 at 20:02
  • $\begingroup$ @STF Yes, you are correct. I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings were not unique. In that case, $c=d$ and the elements of $A,B$ need not be equal. $\endgroup$ Dec 22, 2018 at 22:23
  • $\begingroup$ Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/… $\endgroup$
    – TEX
    Dec 23, 2018 at 9:48
  • $\begingroup$ Sure, I'll give it a try $\endgroup$ Dec 23, 2018 at 9:53

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