<p>If the commodity space is <span class="math-container">$\mathbb{R}^2_+$</span> and the preference is Lexicographic, then with the standard budget sets <span class="math-container">$B=B(p_X, p_Y, M) = \{(x, y) \in \mathbb{R}^2 | p_Xx + p_Yy \leq M\}$</span> where <span class="math-container">$(p_X, p_Y, M) \in\mathbb{R}^3_{++}$</span>
<span class="math-container">$D(p_X, p_Y, M) = \left\{(x, y)\in B| (x, y)\succsim (x',y')  \ \forall (x',y') \in B\right\} = \left\{\left(\frac{M}{p_X}, 0\right)\right\}$</span>
It is non-empty for every closed set <span class="math-container">$B(p_X, p_Y, M)$</span>, and both upper hemi-continuous and lower hemi-continuous.</p>
<p>For <span class="math-container">$\mathbb{R}^2$</span>, consider the following preference relation,
<span class="math-container">\begin{eqnarray*} (x_1, y_1) &amp;\succsim &amp; (x_2, y_2) \\ &amp;\Leftrightarrow &amp; \\ \text{either } (|x_1| &lt; |x_2|) &amp; \text{ or } &amp; (|x_1| = |x_2| \wedge |y_1| \leq |y_2|)\\  \end{eqnarray*}</span>
It is also kind of Lexicographic preference with <span class="math-container">$(0,0)$</span> as the bliss point. If we consider the demand correspondence:
<span class="math-container">$D(B) = \left\{x\in B| x\succsim y  \ \forall y \in B\right\}$</span>
It is also non-empty for every closed set <span class="math-container">$B$</span>, and both upper hemi-continuous and lower hemi-continuous if we consider <span class="math-container">$B = B(p_X, p_Y, M) = \{(x, y) \in \mathbb{R}^2 | p_Xx + p_Yy \leq M\}$</span> where <span class="math-container">$(p_X, p_Y, M) \in\mathbb{R}^3$</span>.</p>

CC BY-SA 4.0
2022-04-30T04:31:27.743Z
You are right. I need explicitly rule-out the trivial case. And for the second example, it is very interesting and I am still checking.
High GPA
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2022-04-30T09:27:16.250Z
But I think the way the problem is proposed, we first need the appropriate topology on the set of subsets of $\mathbb{R}^L$ and then we'll also need the definition of upper and lower hemi-continuity where $D$ is defined on this space of all subsets of $\mathbb{R}^L$ to answer the question.
Amit
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2022-04-30T01:45:09.560Z
2022-04-30T01:55:32.653Z
2022-04-30T01:55:32.653Z
Amit
<p>I think that you should proceed by contradiction assume D is continuous, but <span class="math-container">$\succsim$</span> is not, then for a bundle either the more preferred than or the less preferred than sets are not closed. Choose the problematic set and choose a set B appropriately to get a contradiction of the existence of a maximum (remember that not closed sets might not admit a maximum), this will give you the desired contradiction. Once preferences are continuous, remains to check completeness of preferences to use the classic theorem that if preferences are complete, transitive and continuous, then a utility function representation exists. </p>

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2019-05-26T17:18:20.057Z
2019-05-26T17:18:20.057Z
Regio
<p><span class="math-container">$\succsim$</span> is a weak order over <span class="math-container">$\mathbb R^L$</span>. </p>

<p>For a closed budget set <span class="math-container">$B\subset\mathbb R^L$</span>, define demand correspondence: <span class="math-container">$$D(B)=\{x\in B|x\succsim y\forall y\in B\}$$</span>.</p>

<p>We know that <span class="math-container">$D$</span> is always none empty and both upper and lower hemi-continuous, can we conclude that <span class="math-container">$\succsim$</span> is continuous? Can we conclude that <span class="math-container">$D$</span> is rationalizable by a utility function?</p>

<hr>

<p>My approach: For the first question, I guess we can first show that <span class="math-container">$D$</span> is rationalizable by a utility function <span class="math-container">$u$</span>. Secondly, since <span class="math-container">$D$</span> is continuous, <span class="math-container">$u$</span> must be continuous, and <span class="math-container">$\succsim$</span> is continuous. But I doubt this approach will work, because the second step seems not work.</p>

<p>For the second part, if <span class="math-container">$\succsim$</span> is continuous then it is represented by a continuous utility function. Even if <span class="math-container">$\succsim$</span> is not continous, it must be upper-semi continuous and we can still find a upper semi-continuous utility rationalization for <span class="math-container">$D$</span> by Rader theorem.</p>

<hr>

<p>We could consider a simpler version of the problem: Let the demand function be continuous, is the preference relation also continuous? (This is proved by Mas-Colell in 1978, I think)</p>

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2022-04-30T01:53:19.477Z
What is the domain of $D$? Is it the set of all subsets of $\mathbb{R}^L$? It is important because the definitions of upper hemi-continuity and lower hemi-continuity will be applied accordingly.
Amit
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2022-04-30T02:03:23.810Z
@Amit Hi! If the domain (i.e. the possible budget sets) is restricted to the standard commodity demand budget sets, then I think your results work! In general, if the domain is restricted, then the preference doesn't have to be continuous, or even complete and transitive. However, if we let the domain be all subsets or all compact subsets of $\mathbb R^L$, then it will tell more about the properties of the preference.
High GPA
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2022-04-30T02:14:04.683Z
I believe that the second example I have given in my answer should also pass the test if the domain consist of set of all possible closed sets. If we consider any sequence of closed sets converging to some closed set (we need to define the convergence appropriately), then the sequence of demands will also converge to the demand at the limiting closed set.
Amit
CC BY-SA 4.0
2022-04-30T02:22:07.580Z
Or may be not, I can think of an example. Consider the sequence $B = \{(0,1),(\frac{1}{n},0)\}$, but the demand does not converge to the demand at the limiting budget set.
Amit
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2022-04-30T02:40:52.233Z
I am wondering if there even exists such a preference relation $\succsim$ such that the associated $D$ is always non-empty (for every closed set) and is also both upper and lower hemi-continuous. Of course, if we can show this that there is no such $\succsim$, then the result that $D$ is rationalizable by a continuous $u$ holds trivially.
Amit
CC BY-SA 4.0
2022-04-30T02:57:39.970Z
Ohh yes, I can think of a preference relation $\succsim$, consider $\succsim$ represented by $u(x) = 0$. Associated $D = B$ which is an identity map. It is non-empty, and is both upper hemi-continuous and lower hemi-continuous  (I guess).
Amit
<microeconomics><utility><preferences><decision-theory><rationality>
Demand correspondence is both upper and lower hemi-continuous; is the preference continuous?
High GPA
# Demand correspondence is both upper and lower hemi-continuous; is the preference continuous?

(Asked by: username_0 on 2019-05-24)

$\succsim$ is a weak order over $\mathbb R^L$.

For a closed budget set $B\subset\mathbb R^L$, define demand correspondence: $$D(B)=\{x\in B|x\succsim y\forall y\in B\}$$.

We know that $D$ is always none empty and both upper and lower hemi-continuous, can we conclude that $\succsim$ is continuous? Can we conclude that $D$ is rationalizable by a utility function?

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My approach: For the first question, I guess we can first show that $D$ is rationalizable by a utility function $u$. Secondly, since $D$ is continuous, $u$ must be continuous, and $\succsim$ is continuous. But I doubt this approach will work, because the second step seems not work.

For the second part, if $\succsim$ is continuous then it is represented by a continuous utility function. Even if $\succsim$ is not continous, it must be upper-semi continuous and we can still find a upper semi-continuous utility rationalization for $D$ by Rader theorem.

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We could consider a simpler version of the problem: Let the demand function be continuous, is the preference relation also continuous? (This is proved by Mas-Colell in 1978, I think)

**Tags:** microeconomics, utility, preferences, decision-theory, rationality
> **Comments:**
> **username_1**: What is the domain of $D$? Is it the set of all subsets of $\mathbb{R}^L$? It is important because the definitions of upper hemi-continuity and lower hemi-continuity will be applied accordingly.  
> **username_0**: @Amit Hi! If the domain (i.e. the possible budget sets) is restricted to the standard commodity demand budget sets, then I think your results work! In general, if the domain is restricted, then the preference doesn't have to be continuous, or even complete and transitive. However, if we let the domain be all subsets or all compact subsets of $\mathbb R^L$, then it will tell more about the properties of the preference.  
> **username_1**: I believe that the second example I have given in my answer should also pass the test if the domain consist of set of all possible closed sets. If we consider any sequence of closed sets converging to some closed set (we need to define the convergence appropriately), then the sequence of demands will also converge to the demand at the limiting closed set.  
> **username_1**: Or may be not, I can think of an example. Consider the sequence $B = \{(0,1),(\frac{1}{n},0)\}$, but the demand does not converge to the demand at the limiting budget set.  
> **username_1**: I am wondering if there even exists such a preference relation $\succsim$ such that the associated $D$ is always non-empty (for every closed set) and is also both upper and lower hemi-continuous. Of course, if we can show this that there is no such $\succsim$, then the result that $D$ is rationalizable by a continuous $u$ holds trivially.  
> **username_1**: Ohh yes, I can think of a preference relation $\succsim$, consider $\succsim$ represented by $u(x) = 0$. Associated $D = B$ which is an identity map. It is non-empty, and is both upper hemi-continuous and lower hemi-continuous  (I guess).  

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## 2 answers

### Answer by username_2 on 2019-05-26. Score: 4 | Accepted answer

I think that you should proceed by contradiction assume D is continuous, but $\succsim$ is not, then for a bundle either the more preferred than or the less preferred than sets are not closed. Choose the problematic set and choose a set B appropriately to get a contradiction of the existence of a maximum (remember that not closed sets might not admit a maximum), this will give you the desired contradiction. Once preferences are continuous, remains to check completeness of preferences to use the classic theorem that if preferences are complete, transitive and continuous, then a utility function representation exists.


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### Answer by username_1 on 2022-04-30. Score: 3

If the commodity space is $\mathbb{R}^2\_+$ and the preference is Lexicographic, then with the standard budget sets $B=B(p\_X, p\_Y, M) = \{(x, y) \in \mathbb{R}^2 | p\_Xx + p\_Yy \leq M\}$ where $(p\_X, p\_Y, M) \in\mathbb{R}^3\_{++}$
$D(p\_X, p\_Y, M) = \left\{(x, y)\in B| (x, y)\succsim (x',y') \ \forall (x',y') \in B\right\} = \left\{\left(\frac{M}{p\_X}, 0\right)\right\}$
It is non-empty for every closed set $B(p\_X, p\_Y, M)$, and both upper hemi-continuous and lower hemi-continuous.

For $\mathbb{R}^2$, consider the following preference relation,
\begin{eqnarray\*} (x\_1, y\_1) &\succsim & (x\_2, y\_2) \\ &\Leftrightarrow & \\ \text{either } (|x\_1| < |x\_2|) & \text{ or } & (|x\_1| = |x\_2| \wedge |y\_1| \leq |y\_2|)\\ \end{eqnarray\*}
It is also kind of Lexicographic preference with $(0,0)$ as the bliss point. If we consider the demand correspondence:
$D(B) = \left\{x\in B| x\succsim y \ \forall y \in B\right\}$
It is also non-empty for every closed set $B$, and both upper hemi-continuous and lower hemi-continuous if we consider $B = B(p\_X, p\_Y, M) = \{(x, y) \in \mathbb{R}^2 | p\_Xx + p\_Yy \leq M\}$ where $(p\_X, p\_Y, M) \in\mathbb{R}^3$.

> **Comments:**
> **username_0**: You are right. I need explicitly rule-out the trivial case. And for the second example, it is very interesting and I am still checking.  
> **username_1**: But I think the way the problem is proposed, we first need the appropriate topology on the set of subsets of $\mathbb{R}^L$ and then we'll also need the definition of upper and lower hemi-continuity where $D$ is defined on this space of all subsets of $\mathbb{R}^L$ to answer the question.  

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