<p>Continuous functions attain a minimum and a maximum value over a compact (closed and bounded set). This is a well known theorem.</p>
<p>Since the utility function <span class="math-container">$u_i$</span> representing <span class="math-container">$\succeq_i$</span> is continuous and <span class="math-container">$X_i$</span> is closed and bounded, then <span class="math-container">$u_i$</span> must attain a maximum value at some point <span class="math-container">$x = (x_1,\dots,x_L)$</span> in <span class="math-container">$X_i$</span>.</p>
<p>Since <span class="math-container">$u_i$</span> has a maximum value at <span class="math-container">$x$</span>, there is no <span class="math-container">$y \in X_i : y \succ_i x$</span>.</p>
<p>Therefore, <span class="math-container">$\succeq_i$</span> is not locally non-satiable.</p>
<p>The definition of locally non-satiable I use is <span class="math-container">$\forall x \in X \forall \epsilon &gt; 0 \exists y \in X : ||y-x|| \leq \epsilon$</span> and <span class="math-container">$y \succ x$</span>.</p>
<p>Note this definition implies the non-existence of maximum values for the corresponding utility function, if the preferences are indeed representable by a utility function (which happens when the preferences are complete, transitive and continuous).</p>

CC BY-SA 4.0
2023-02-14T21:43:39.433Z
I think that you are wrong, it should be, there is no $y$ in the consumption set preferred to $x$. Is the answer so simple?
Maximilian
CC BY-SA 4.0
2023-02-14T21:52:43.250Z
Yes, it is a simple application of Weierestrass’s theorem. Since $u_i$ attains its maximum vale on $x$, there is no point $y$ strictly preferred to $x$.
Nicolas Torres
CC BY-SA 4.0
2023-02-14T21:55:24.030Z
Right, I was pointing out this just because you wrote at least as good as instead of strictly preferred. Thanks
Maximilian
CC BY-SA 4.0
2023-02-14T21:57:06.390Z
@Maximilian I just corrected it as you pointed that out, it was actually a typo.
Nicolas Torres
CC BY-SA 4.0
2023-02-14T21:37:49.900Z
2023-02-14T21:55:59.513Z
2023-02-14T21:55:59.513Z
Nicolas Torres
<p>I'm studying advanced micro from the Mas-Colell book (exercise 16.C.1)</p>
<p>I was wondering if anyone can help me to solve the following exercise. I have no idea how to deal with it</p>
<p>Show that if a consumption set <span class="math-container">$X_i \subset \mathbb{R}^{L}$</span> is nonempty, closed, and bounded and the preference relation <span class="math-container">$\succeq_i$</span> on <span class="math-container">$X_i$</span> is continuous, then <span class="math-container">$\succeq_i$</span> cannot be locally nonsatiated. [Hint: Show that the continuous utility function representing <span class="math-container">$\succeq_i$</span> must have a maximum on <span class="math-container">$X_i$</span>]</p>

CC BY-SA 4.0
2023-02-14T21:37:05.977Z
Do you have a problem showing that a representing continuous utility function has a maximum? Or do you have a problem relating the existence of such a maximum to local nonsatiation?
Michael Greinecker
<microeconomics><preferences><self-study>
Challenging question in microeconomics - local nonsatiation
Maximilian
# Challenging question in microeconomics - local nonsatiation

(Asked by: username_0 on 2023-02-14)

I'm studying advanced micro from the Mas-Colell book (exercise 16.C.1)

I was wondering if anyone can help me to solve the following exercise. I have no idea how to deal with it

Show that if a consumption set $X\_i \subset \mathbb{R}^{L}$ is nonempty, closed, and bounded and the preference relation $\succeq\_i$ on $X\_i$ is continuous, then $\succeq\_i$ cannot be locally nonsatiated. [Hint: Show that the continuous utility function representing $\succeq\_i$ must have a maximum on $X\_i$]

**Tags:** microeconomics, preferences, self-study

> **Comments:**
> **username_1**: Do you have a problem showing that a representing continuous utility function has a maximum? Or do you have a problem relating the existence of such a maximum to local nonsatiation?  

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## 1 answers

### Answer by username_2 on 2023-02-14. Score: 2 | Accepted answer

Continuous functions attain a minimum and a maximum value over a compact (closed and bounded set). This is a well known theorem.

Since the utility function $u\_i$ representing $\succeq\_i$ is continuous and $X\_i$ is closed and bounded, then $u\_i$ must attain a maximum value at some point $x = (x\_1,\dots,x\_L)$ in $X\_i$.

Since $u\_i$ has a maximum value at $x$, there is no $y \in X\_i : y \succ\_i x$.

Therefore, $\succeq\_i$ is not locally non-satiable.

The definition of locally non-satiable I use is $\forall x \in X \forall \epsilon > 0 \exists y \in X : ||y-x|| \leq \epsilon$ and $y \succ x$.

Note this definition implies the non-existence of maximum values for the corresponding utility function, if the preferences are indeed representable by a utility function (which happens when the preferences are complete, transitive and continuous).

> **Comments:**
> **username_0**: I think that you are wrong, it should be, there is no $y$ in the consumption set preferred to $x$. Is the answer so simple?  
> **username_2**: Yes, it is a simple application of Weierestrass’s theorem. Since $u_i$ attains its maximum vale on $x$, there is no point $y$ strictly preferred to $x$.  
> **username_0**: Right, I was pointing out this just because you wrote at least as good as instead of strictly preferred. Thanks  
> **username_2**: @username_0 I just corrected it as you pointed that out, it was actually a typo.  

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