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

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

Consider other four real numbers $c,d,e,f$ each in $[0,1]$, all different between each other.

Assume there exists a way of ordering the four numbers in each set $A,B$ such that $$ \begin{cases} w^A_1+w^B_1=c\\ w^A_2+w^B_2=d\\ w^A_3+w^B_3=e\\ w^A_4+w^B_4=f\\ \end{cases} $$ where

  • $w^A_h$ denotes the $h$th element of set $A$ once we have ordered its 4 elements

  • $w^B_h$ denotes the $h$th element of set $B$ once we have ordered its 4 elements

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

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

(similar question here but with 2 elements per set)


Maybe the claim is wrong? Let $\{a_1,a_2,a_3,a_4\}$ be the elements of $A$ and $\{b_1,b_2,b_3,b_4\}$ be the elements of $B$. We could have:

order I $$ \begin{cases} a_2+b_3=c\\ a_4+b_4=d\\ a_1+b_1=e\\ a_3+b_2=f \end{cases} $$

and

order II $$ \begin{cases} a_1+b_2=c\\ a_2+b_1=d\\ a_3+b_4=e\\ a_4+b_3=f \end{cases} $$

which implies $$ \begin{cases} a_2+b_3=a_1+b_2\\ a_4+b_4=a_2+b_1\\ a_1+b_1=a_3+b_4\\ a_3+b_2=a_4+b_3 \end{cases} $$ Does this imply that two numbers in $A$ are equal and/or that two numbers in $B$ are equal?

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The claim is wrong. Pick the following partial solution of the last system: $(a_1,a_2,a_3,a_4)=(0,a,a’,a+a’)$ and $(b_1,b_2,b_3,b_4)=(a’,a+b,b,0)$. Then $(c,d,e,f)=(a+b,a+a’,a’,a+a’+b)$. For instance, we can put $a=0.1$, $a’=0.15$ and $b=0.2$. Then $(a_1,a_2,a_3,a_4)=(0,0.1,0.15,0.25)$, $(b_1,b_2,b_3,b_4)=(0.15,0.3,0.2,0)$, and $(c,d,e,f)=(0.3,0.25,0.15,0.45)$.

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