<p>You just need to use the violation of transitivity and proceed by contradiction. </p>

<p>Suppose you have  that <span class="math-container">$(x_1,y_1)R(x_2,y_2)$</span> and <span class="math-container">$(x_2,y_2)R(x_3,y_3)$</span> and a utility function, <span class="math-container">$u:[-1,1]^2\rightarrow \mathbb{R}$</span>, exists, then <span class="math-container">$u(x_1,y_1)\geq u(x_2,y_2)$</span> (these are two reals) and <span class="math-container">$u(x_2,y_2)\geq u(x_3,y_3)$</span>. (another two reals). Since the reals are transitive, we conclude  <span class="math-container">$u(x_1,y_1)\geq u(x_3,y_3)$</span> which in turn implies that <span class="math-container">$(x_1,y_1)R(x_3,y_3)$</span>. However this is a contradiction (if you choose carefully your three bunddles). </p>

CC BY-SA 4.0
2019-07-03T00:18:59.450Z
2019-07-03T00:18:59.450Z
Regio
<p>Transitivity and completeness are actually <em>necessary</em> for the existence of a utility representation. Whenever you prove that preferences fail to be complete or transitive you can conclude that they do <em>not</em> admit a utility function. 
For finite choice sets <span class="math-container">$X$</span>, transitivity and completeness are necessary and sufficient, see Theorem 5 <a href="https://pdfs.semanticscholar.org/1cdc/4f88148c2573769f2763897f831963015524.pdf" rel="nofollow noreferrer">here</a>.</p>

<p>Contrary to what you suggest, continuity is not itself necessary, but it is almost necessary. What you need is a condition called separability. Say that <span class="math-container">$\succcurlyeq$</span> is separable if there exists a <em>countable</em> set <span class="math-container">$Z \subseteq X$</span> such that for every <span class="math-container">$x,y\in X$</span> there exists some <span class="math-container">$z\in Z$</span> such that <span class="math-container">$x\succcurlyeq z \succcurlyeq y$</span>. </p>

<p><strong>Theorem</strong> A preference relation <span class="math-container">$\succcurlyeq$</span> on <span class="math-container">$X$</span> admits a utility representation if and only if it is complete, transitive, and separable.</p>

<p>This is actually a very old result by Cantor that precedes the well-known result by Debreu&mdash;which assumes continuity. Cantor's result was brought into Economics by Kreps. You can find a proof <a href="https://pdfs.semanticscholar.org/1cdc/4f88148c2573769f2763897f831963015524.pdf" rel="nofollow noreferrer">here, Theorem 9</a>.</p>

CC BY-SA 4.0
2019-08-06T17:12:02.083Z
"Whenever you prove that preferences fail to be *continuous* or transitive..." Did you mean "**complete** or transitive"?
Herr K.
CC BY-SA 4.0
2019-08-06T18:20:52.270Z
Let me suggest the edit above.
Bayesian
CC BY-SA 4.0
2019-08-06T01:33:06.737Z
2019-08-06T18:19:37.657Z
2019-08-06T18:19:37.657Z
brunosalcedo
<p><strong>Setting:</strong> We have two choices of goods <span class="math-container">$(x_1,y_1)$</span> and <span class="math-container">$(x_2,y_2)$</span> from the set of choices <span class="math-container">$[-1,1]^2$</span>. Moreover, we have the following preference relation <span class="math-container">$$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff |x_1|\geq|x_2|\&gt;\&gt;\text{or}\&gt;\&gt; |y_1|\geq|y_2|$$</span></p>

<p><strong>Question:</strong> We have to check if there exists a utility function reprensation of this preference relation.</p>

<p><strong>My attempt:</strong> So from what I have learned, we know that a preference relation admits a utility function representation <em>if</em> it is rational (reflexive, complete, transitive) and continuous. I have found that this preference relation is not transitive, but this does not mean that there does not exist a utility function representation, because the aforementioned statement is not an <em>if and only if</em> statement.</p>

<p>Moreover, I thought we could try to derive a contradiction from the fact that if there exists a utility function <span class="math-container">$u$</span> representation of the preference relation, then we have <span class="math-container">$$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff u(x_1,y_1)\geq u(x_2,y_2)$$</span>
I tried to use the fact that the relation is not transitive to derive a contradiction by using the statement above, but was unsuccessful.</p>

<p>Sadly, these are the two main theorems/propositions that I've learned to solve these problems.</p>

<p>Any help is appreciated!</p>

<utility><preferences>
Showing that a preference relation admits a utility function representation
Charlie Shuffler
# Showing that a preference relation admits a utility function representation

(Asked by: username_0 on 2019-07-02)

**Setting:** We have two choices of goods $(x\_1,y\_1)$ and $(x\_2,y\_2)$ from the set of choices $[-1,1]^2$. Moreover, we have the following preference relation $$(x\_1,y\_1)\mathcal{R}(x\_2,y\_2)\iff |x\_1|\geq|x\_2|\>\>\text{or}\>\> |y\_1|\geq|y\_2|$$

**Question:** We have to check if there exists a utility function reprensation of this preference relation.

**My attempt:** So from what I have learned, we know that a preference relation admits a utility function representation *if* it is rational (reflexive, complete, transitive) and continuous. I have found that this preference relation is not transitive, but this does not mean that there does not exist a utility function representation, because the aforementioned statement is not an *if and only if* statement.

Moreover, I thought we could try to derive a contradiction from the fact that if there exists a utility function $u$ representation of the preference relation, then we have $$(x\_1,y\_1)\mathcal{R}(x\_2,y\_2)\iff u(x\_1,y\_1)\geq u(x\_2,y\_2)$$
I tried to use the fact that the relation is not transitive to derive a contradiction by using the statement above, but was unsuccessful.

Sadly, these are the two main theorems/propositions that I've learned to solve these problems.

Any help is appreciated!

**Tags:** utility, preferences

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## 2 answers

### Answer by username_1 on 2019-07-03. Score: 3

You just need to use the violation of transitivity and proceed by contradiction.

Suppose you have that $(x\_1,y\_1)R(x\_2,y\_2)$ and $(x\_2,y\_2)R(x\_3,y\_3)$ and a utility function, $u:[-1,1]^2\rightarrow \mathbb{R}$, exists, then $u(x\_1,y\_1)\geq u(x\_2,y\_2)$ (these are two reals) and $u(x\_2,y\_2)\geq u(x\_3,y\_3)$. (another two reals). Since the reals are transitive, we conclude $u(x\_1,y\_1)\geq u(x\_3,y\_3)$ which in turn implies that $(x\_1,y\_1)R(x\_3,y\_3)$. However this is a contradiction (if you choose carefully your three bunddles).


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### Answer by username_2 on 2019-08-06. Score: 2

Transitivity and completeness are actually *necessary* for the existence of a utility representation. Whenever you prove that preferences fail to be complete or transitive you can conclude that they do *not* admit a utility function.
For finite choice sets $X$, transitivity and completeness are necessary and sufficient, see Theorem 5 [here](https://pdfs.semanticscholar.org/1cdc/4f88148c2573769f2763897f831963015524.pdf).

Contrary to what you suggest, continuity is not itself necessary, but it is almost necessary. What you need is a condition called separability. Say that $\succcurlyeq$ is separable if there exists a *countable* set $Z \subseteq X$ such that for every $x,y\in X$ there exists some $z\in Z$ such that $x\succcurlyeq z \succcurlyeq y$.

**Theorem** A preference relation $\succcurlyeq$ on $X$ admits a utility representation if and only if it is complete, transitive, and separable.

This is actually a very old result by Cantor that precedes the well-known result by Debreu—which assumes continuity. Cantor's result was brought into Economics by Kreps. You can find a proof [here, Theorem 9](https://pdfs.semanticscholar.org/1cdc/4f88148c2573769f2763897f831963015524.pdf).

> **Comments:**
> **username_3**: "Whenever you prove that preferences fail to be *continuous* or transitive..." Did you mean "**complete** or transitive"?  
> **username_4**: Let me suggest the edit above.  

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