Implications of multiple ways to order four numbers

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?

• 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. Dec 22, 2018 at 19:14
• Yes, sorry, I have deleted the summing up to one. Thanks for the observation.
– TEX
Dec 22, 2018 at 19:30

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.

• 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).
– TEX
Dec 22, 2018 at 20:02
• Is it fair to say that if $c\neq d$ then your proof is correct?
– TEX
Dec 22, 2018 at 20:02
• @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. Dec 22, 2018 at 22:23
• Could you help with 4 elements per set if you have some time? math.stackexchange.com/questions/3049836/…
– TEX
Dec 23, 2018 at 9:48
• Sure, I'll give it a try Dec 23, 2018 at 9:53