<p>Το take this one out of the Unanswered queue:</p>

<p>The completeness property of the preference relation, implies that <strong>for all non-negativ</strong>e $t$ we will be able to form and declare the preference relation. Hence</p>

<p>$$A \cup B = \mathbb{R_+}$$</p>

<p>By monotonicity we have $B=[0, \overline{t}],\;\; A=[\underline{t}, \infty)$.</p>

<p>Ad absurdum, assume that $\overline{t} &lt; \underline{t}$. Then there exists an open interval $(\overline{t},\underline{t})$ that does not belong to the union of $A$ and $B$. But then we arrive at $A \cup B \neq \mathbb{R_+}$ which contradicts the implications of the completeness property. </p>

<p>So we conclude that $\overline{t} \geq \underline{t}$, which implies that the intersection $A \cap B$ is non-empty (even if it may contain a single element, when $\overline{t} = \underline{t}$).</p>

CC BY-SA 3.0
2017-06-16T12:19:04.730Z
2017-06-16T12:19:04.730Z
Alecos Papadopoulos
<p>In Advanced Microeconomic theory by Jehle and Reny there is a proof of the theorem which states the existence of utility function.</p>

<p>In order to prove the existence of a utility function $u(\mathbf{x})$ which represents binary relation $\succeq$ if it is complete, transitive, continuous, and strictly monotonic, it is suggested to consider a mapping</p>

<p>$u: \mathbb{R_+^n} \to \mathbb{R}$  such that $u(\mathbf{x})e\sim \mathbf{x}$ is satisfied, where $\mathbf{x}$ is a bundle, $u(\mathbf{x})$ is some number and $\mathbf{e}$ is a bundle which contains one of every good.</p>

<p>So <strong>first</strong> we need to show that there always exists such number $u(\mathbf{x})$. 
To do this consider two sets:</p>

<p>$A \equiv \{t \geq 0 \mid t\mathbf{e} \succeq \mathbf{x}\}$</p>

<p>$B \equiv \{t \geq 0 \mid t\mathbf{e} \preceq \mathbf{x}\}$</p>

<p>if $t^* \in  A \cap B$, then $t^*\mathbf{e} \sim \mathbf{x}$, so we need to show that $A \cap B$ is nonempty.</p>

<p>The continuity of $\succeq$ implies that both A and B are closed in $\mathbb{R_+}$. By strict monotonicity, $t \in A$ implies $t' \in A,$ $\forall$  $ t'\geq t$. So $A=[\underline{t}, \infty)$. 
Similarly $B=[0, \overline{t}]$</p>

<p>For any $t \geq 0$, completeness of $\succeq$ implies that either $t\mathbf{e} \succeq \mathbf{x}$ or $t\mathbf{e} \preceq \mathbf{x}$, 
i.e., $t \in A \cup B$</p>

<p>$\mathbb{R_+} = A \cup B = [0, \overline{t}] \cup [\underline{t}, \infty)$.</p>

<p>So in order to $A \cap B$ be nonempty it should be that $\underline{t} \leq \overline{t}$.</p>

<p>But is it always like this? I can't see why the last inequality should hold always.</p>

CC BY-SA 3.0
2016-10-26T03:59:35.770Z
I think you pretty much finished it. If $\underline{t} > \overline{t}$, then $A \cup B$ does not contain any numbers in the (nonempty) interval $(\overline{t}, \underline{t})$. But this contradicts the fact you already showed that $A \cup B = \mathbb{R}_+$. So we must have $\underline{t} \leq \overline{t}$.
usul
CC BY-SA 3.0
2017-04-13T21:16:29.737Z
Just one further reminder, this is far from the only axiomization of the utility function.  You are complete.
Dave Harris
<microeconomics><mathematical-economics><utility>
Proving existence of utility function
Vlad Lev
# Proving existence of utility function

(Asked by: username_0 on 2016-10-25)

In Advanced Microeconomic theory by Jehle and Reny there is a proof of the theorem which states the existence of utility function.

In order to prove the existence of a utility function $u(\mathbf{x})$ which represents binary relation $\succeq$ if it is complete, transitive, continuous, and strictly monotonic, it is suggested to consider a mapping

$u: \mathbb{R\_+^n} \to \mathbb{R}$ such that $u(\mathbf{x})e\sim \mathbf{x}$ is satisfied, where $\mathbf{x}$ is a bundle, $u(\mathbf{x})$ is some number and $\mathbf{e}$ is a bundle which contains one of every good.

So **first** we need to show that there always exists such number $u(\mathbf{x})$.
To do this consider two sets:

$A \equiv \{t \geq 0 \mid t\mathbf{e} \succeq \mathbf{x}\}$

$B \equiv \{t \geq 0 \mid t\mathbf{e} \preceq \mathbf{x}\}$

if $t^\* \in A \cap B$, then $t^\*\mathbf{e} \sim \mathbf{x}$, so we need to show that $A \cap B$ is nonempty.

The continuity of $\succeq$ implies that both A and B are closed in $\mathbb{R\_+}$. By strict monotonicity, $t \in A$ implies $t' \in A,$ $\forall$ $ t'\geq t$. So $A=[\underline{t}, \infty)$.
Similarly $B=[0, \overline{t}]$

For any $t \geq 0$, completeness of $\succeq$ implies that either $t\mathbf{e} \succeq \mathbf{x}$ or $t\mathbf{e} \preceq \mathbf{x}$,
i.e., $t \in A \cup B$

$\mathbb{R\_+} = A \cup B = [0, \overline{t}] \cup [\underline{t}, \infty)$.

So in order to $A \cap B$ be nonempty it should be that $\underline{t} \leq \overline{t}$.

But is it always like this? I can't see why the last inequality should hold always.

**Tags:** microeconomics, mathematical-economics, utility
> **Comments:**
> **username_1**: I think you pretty much finished it. If $\underline{t} > \overline{t}$, then $A \cup B$ does not contain any numbers in the (nonempty) interval $(\overline{t}, \underline{t})$. But this contradicts the fact you already showed that $A \cup B = \mathbb{R}_+$. So we must have $\underline{t} \leq \overline{t}$.  
> **username_2**: Just one further reminder, this is far from the only axiomization of the utility function.  You are complete.  

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## 1 answers

### Answer by username_3 on 2017-06-16. Score: 2 | Accepted answer

Το take this one out of the Unanswered queue:

The completeness property of the preference relation, implies that **for all non-negativ**e $t$ we will be able to form and declare the preference relation. Hence

$$A \cup B = \mathbb{R\_+}$$

By monotonicity we have $B=[0, \overline{t}],\;\; A=[\underline{t}, \infty)$.

Ad absurdum, assume that $\overline{t} < \underline{t}$. Then there exists an open interval $(\overline{t},\underline{t})$ that does not belong to the union of $A$ and $B$. But then we arrive at $A \cup B \neq \mathbb{R\_+}$ which contradicts the implications of the completeness property.

So we conclude that $\overline{t} \geq \underline{t}$, which implies that the intersection $A \cap B$ is non-empty (even if it may contain a single element, when $\overline{t} = \underline{t}$).


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