5
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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 $\underline{t} \leq \overline{t}$. $\endgroup$ – usul Oct 26 '16 at 3:59
  • $\begingroup$ Just one further reminder, this is far from the only axiomization of the utility function. You are complete. $\endgroup$ – Dave Harris Apr 13 '17 at 21:16
2
$\begingroup$

Το take this one out of the Unanswered queue:

The completeness property of the preference relation, implies that for all non-negative $t$ 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 $\overline{t} = \underline{t}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.