<p>My textbook, <strong><em>Microeconomic Theory</em></strong> by <em>Mas-Colell, Whinston, and Green</em> states that given a preference relation <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span>, </p>

<blockquote>
  <p><strong>Strict preference relation</strong> <span class="math-container">$\succ$</span> is defined by 
  <span class="math-container">$ x \succ y \iff x \succsim y \text{ but not } y \succsim x $</span></p>
  
  <p><strong>Indifference relation</strong> <span class="math-container">$\sim$</span> is defined by 
  <span class="math-container">$ x \sim y \iff x \succsim y \text{ and } y \succsim x $</span></p>
  
  <p><strong>Completeness property</strong>: For all <span class="math-container">$x,y \in X$</span>, we have either <span class="math-container">$x \succsim y$</span> or <span class="math-container">$y \succsim x$</span> or both.</p>
</blockquote>

<p>Suppose <span class="math-container">$\succsim$</span> on <span class="math-container">$X$</span> is complete. Define <span class="math-container">$A$</span> to be the truth set of <span class="math-container">$x \succsim y$</span>, <span class="math-container">$B$</span> to be the truth set of <span class="math-container">$y \succsim x$</span>. </p>

<p>Since <span class="math-container">$\succsim$</span> is complete, <span class="math-container">$A \cup B$</span> is the universal set. </p>

<p>First, <span class="math-container">$x \succ y \iff x \succsim y \land \lnot (y \succsim x)$</span>, hence the truth set of <span class="math-container">$x \succ y$</span> is <span class="math-container">$A \setminus B$</span>. </p>

<p>Next, <span class="math-container">$y \succ x \iff y \succsim x \land \lnot (x \succsim y)$</span>, hence the truth set of <span class="math-container">$y \succ x$</span> is <span class="math-container">$B \setminus A$</span>.</p>

<p>Finally, <span class="math-container">$x \sim y \iff x \succsim y \land y \succsim x$</span>, hence the truth set of <span class="math-container">$x \sim y$</span> is <span class="math-container">$A \cap B$</span>.</p>

<p>Using the identity <span class="math-container">$A \cup B = (A \setminus B) \sqcup (A \cap B) \sqcup (B \setminus A)$</span>, is it safe to say that </p>

<blockquote>
  <p>If <span class="math-container">$\succsim$</span> is complete, then for any <span class="math-container">$x,y \in X$</span> exactly one of <span class="math-container">$x \succ y$</span> or <span class="math-container">$x \sim y$</span> or <span class="math-container">$y \succ x$</span> holds. </p>
</blockquote>

CC BY-SA 4.0
2019-02-01T07:02:14.690Z
Yes, that seems perfectly valid reasoning to me
J. Dekker
CC BY-SA 4.0
2019-02-01T22:26:07.437Z
This just applies the definitions of completeness and relations together. You don't need set theory to show this. You could just work from the mathematical definitions. Anyway...yeah...it is true.
123
<preferences>
Completeness Axiom in Preference Relation
user20961
# Completeness Axiom in Preference Relation

(Asked by: username_0 on 2019-02-01)

My textbook, ***Microeconomic Theory*** by *Mas-Colell, Whinston, and Green* states that given a preference relation $\succsim$ on $X$,

> **Strict preference relation** $\succ$ is defined by
> $ x \succ y \iff x \succsim y \text{ but not } y \succsim x $
>
> **Indifference relation** $\sim$ is defined by
> $ x \sim y \iff x \succsim y \text{ and } y \succsim x $
>
> **Completeness property**: For all $x,y \in X$, we have either $x \succsim y$ or $y \succsim x$ or both.

Suppose $\succsim$ on $X$ is complete. Define $A$ to be the truth set of $x \succsim y$, $B$ to be the truth set of $y \succsim x$.

Since $\succsim$ is complete, $A \cup B$ is the universal set.

First, $x \succ y \iff x \succsim y \land \lnot (y \succsim x)$, hence the truth set of $x \succ y$ is $A \setminus B$.

Next, $y \succ x \iff y \succsim x \land \lnot (x \succsim y)$, hence the truth set of $y \succ x$ is $B \setminus A$.

Finally, $x \sim y \iff x \succsim y \land y \succsim x$, hence the truth set of $x \sim y$ is $A \cap B$.

Using the identity $A \cup B = (A \setminus B) \sqcup (A \cap B) \sqcup (B \setminus A)$, is it safe to say that

> If $\succsim$ is complete, then for any $x,y \in X$ exactly one of $x \succ y$ or $x \sim y$ or $y \succ x$ holds.

**Tags:** preferences
> **Comments:**
> **username_1**: Yes, that seems perfectly valid reasoning to me  
> **username_2**: This just applies the definitions of completeness and relations together. You don't need set theory to show this. You could just work from the mathematical definitions. Anyway...yeah...it is true.  

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## 0 answers

