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y is weakly preferred over x if and only if x+y ≤ 4 defines a preference relation on {0,1,2,3}. True or False?. I can see why it's not transitive but I was told it was incomplete if we take 2 and 3. If the bundle violates the condition x+y ≤ 4 (x ≤ 4-y) doesn't that just means x is weakly preferred over y hence complete.

Just like we deal with y is weakly preferred over x iff x≤y on the set X={0,1,2}.

Many thanks,

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Define the Set of Alternatives X as X ={0,1,2,3}

Define the weakly preferred relation on X as

$\succsim : X \to X$ such that $(\forall x \in X)(\forall y \in X)$ , it is the case that

$y \succsim x \iff x+y \leq 4$

The definition for completeness can be referred below:- $(\forall x \in X) (\forall y \in X) [ x \neq y \Rightarrow x \succsim y \: \lor y \succsim x] \Rightarrow \: \succsim $ is complete

Clearly, we can find counter example to show that $\succsim$ is not complete.

Notice from the definition of $\succsim$, it also holds that $\neg (y \succsim x) \iff \neg(x+y \leq 4)$

Counterexample can be, $\neg(3 \succsim 2)$ as $2+3 > 4$ and $\neg(2 \succsim 3) $ as $3+2>4$

Since $(\exists x \in X)(\exists y \in X) [x \neq y \land \neg (x \succsim y) \land \neg(y \succsim x)]$ is true ,which is negation of definiton of completeness, it can be concluded that $\succsim$ is not complete.

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