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Question: Let $N=\{1,2,3\}$ be a set of three people and $X=\{a,b,c,d\}$ a set of 4 political parties. The individuals have the following preferences:

Individual 1: $a \succ_1 b \sim_1 c \succ_1 d$

Individual 2: $d \sim_2 a \succ_2 c \succ_2 b$

Individual 3: $a \succ_3 b \succ_3 c \succ_3 d$

List all pairs of $(x,y) \in X \times X$ such that $x$ Pareto dominates $y$ and deduce the optimal Pareto set.

I think I am being thrown off by the $X \times X$ bit. But it seems like if I multiply out $X$ I will get the following:

$X \times X = \{(a,a),(a,b),(a,c),(a,d),(b,b),(b,c),(b,d),(c,c),(c,d),(d,d)\}$

So if I define the Pareto outcomes as $x \succeq^P y \iff x \succeq_i y$ for all $i=1,\dots,n$. Then my answer to the question would be the pairs are:

$(a,b),(a,c)$ since those are the only two combinations which dominate in all three individuals. I don't think it makes sense to examine the $(a,a),(b,b),...$ case, but I could be completely wrong. For some reason I can't seem to wrap my head around it, and I could have gone wrong right from the beginning.

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I am somewhat unsure where your exact difficulty lies (and hence what your question is). As you write the two Pareto impromevents are $(a,b),(a,c)$. Meaning $b$ and $c$ can be improved upon and are not efficient. Also $(a,d)$ is an improvement, notice that individual 2 is indifferent while others prefer $a$ to $d$. Alternative $a$ cannot be improved upon, and hence it is Pareto-efficient. So the Pareto-set would be $\left\{a\right\}$.

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