<p>Define the Set of Alternatives X as
X ={0,1,2,3}</p>
<p>Define the weakly preferred relation on X as</p>
<p><span class="math-container">$\succsim : X \to X$</span> such that <span class="math-container">$(\forall x \in X)(\forall y \in X)$</span> , it is the case that</p>
<p><span class="math-container">$y \succsim x \iff x+y \leq 4$</span></p>
<p>The definition for completeness can be referred below:-
<span class="math-container">$(\forall x \in X) (\forall y \in X) [ x \neq y \Rightarrow x \succsim y \: \lor y \succsim x] \Rightarrow \: \succsim $</span> is complete</p>
<p>Clearly, we can find counter example to show that <span class="math-container">$\succsim$</span> is not complete.</p>
<p>Notice from the definition of <span class="math-container">$\succsim$</span>, it also holds that
<span class="math-container">$\neg (y \succsim x) \iff \neg(x+y \leq 4)$</span></p>
<p>Counterexample can be, <span class="math-container">$\neg(3 \succsim 2)$</span> as <span class="math-container">$2+3 &gt; 4$</span> and <span class="math-container">$\neg(2 \succsim 3) $</span> as <span class="math-container">$3+2&gt;4$</span></p>
<p>Since <span class="math-container">$(\exists x \in X)(\exists y \in X) [x \neq y \land \neg (x \succsim y) \land \neg(y \succsim x)]$</span> is true ,which is negation of definiton of completeness, it can be concluded that <span class="math-container">$\succsim$</span> is not complete.</p>

CC BY-SA 4.0
2023-10-11T02:44:40.147Z
2023-10-11T05:11:18.603Z
2023-10-11T05:11:18.603Z
Divyam Verma
<p>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.</p>
<p>Just like we deal with y is weakly preferred over x iff x≤y on the set X={0,1,2}.</p>
<p>Many thanks,</p>

<microeconomics><preferences>
y is weakly preferred over x if and only if x+y ≤ 4 defines a preference relation on {0,1,2,3} why is this incomplete?
Froter
# y is weakly preferred over x if and only if x+y ≤ 4 defines a preference relation on {0,1,2,3} why is this incomplete?

(Asked by: username_0 on 2023-10-10)

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,

**Tags:** microeconomics, preferences


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## 1 answers

### Answer by username_1 on 2023-10-11. Score: 2

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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