<p>Great question, never thought about this: </p>

<p>So the main thing that LNS preferences get rid of is a satiation point, preventing the budget from binding in the consumer's problem. So there are tons of preferences that are non-monotonic, strictly convex, with a satiation point. </p>

<p>Now a simple example of weakly convex with thick IC is indfference everywhere. So $\forall x,y\in X, x\sim y$. In this case the IC is all of $\mathbb{R}$. </p>

<p>I agree with you on that strict convexity eliminates thick IC. I'm working on a proof but defining thick IC in mathmatical terms is tricky. However, just recall non-thick IC is not enough to guarantee LNS, so that's why strict convexity is not a substitute for LNS in major theorems like the 2nd Welfare Theorem. </p>

CC BY-SA 3.0
2016-08-21T06:04:17.090Z
Thanks for the answer! Indifference everywhere is a great example of thick IC which satisfy weak convexity. I agree with you that while strong convexity may get rid of thick IC (proof pending, im trying to work out one myself too), they dont guarantee LNS, and the usefullness for some theorems vanishes. Nevertheless, I think that the proof would be of value by itslef, because it would asure that LNS is not the only property that can eliminate thickness, aspect that is not mentioned at all in micro textbooks, it just broadens our understanding of different axioms to the shape of u(.) functions
thekiciminister
CC BY-SA 3.0
2016-08-20T12:51:15.763Z
2016-08-20T12:51:15.763Z
VCG
<p>My approach would be to define thick indifference curves in terms of a stronger form of local non-satiation:</p>

<blockquote>
  <p><strong>Definition (thick indifference curves)</strong> Preferences are said to have thick indifference curves if there exists at least one bundle <span class="math-container">$A\in\mathbb{R}^l$</span> and an open ball
  <span class="math-container">$\mathscr{B}(A)$</span> around <span class="math-container">$A$</span> such that <span class="math-container">$A'\sim A$</span> for <em>every</em>
  <span class="math-container">$A'\in\mathscr{B}(A)$</span>.</p>
</blockquote>

<p>Let's set up a definition for strict convexity to be sure we are on the same page:</p>

<blockquote>
  <p><strong>Definition (strict convexity)</strong> Preferences obey strict convexity if</p>
  
  <p><span class="math-container">$$A\sim B\implies \lambda A+(1-\lambda) B\succ A\sim B$$</span></p>
  
  <p>for <span class="math-container">$A,B\in\mathbb{R}^l$</span> and any <span class="math-container">$\lambda\in(0,1)$</span>.</p>
</blockquote>

<p>Now we can state the desired result:</p>

<blockquote>
  <p><strong>Proposition</strong> If <span class="math-container">$\succsim$</span> is strictly convex then <span class="math-container">$\succsim$</span> does not have thick indifference curves.</p>
</blockquote>

<p><strong>Proof</strong> Suppose that <span class="math-container">$\succsim$</span> has thick indifference curves. Then there exists an <span class="math-container">$A$</span> such that <span class="math-container">$A'\sim A$</span> for all <span class="math-container">$A'\in\mathscr{B}(A)$</span>. Thus, fix two bundles <span class="math-container">$A$</span> and <span class="math-container">$B$</span> such that <span class="math-container">$A\sim B$</span> and <span class="math-container">$B\in\mathscr{B}(A)$</span>. We know that 
<span class="math-container">$$\lambda A+(1-\lambda) B \in \mathscr{B}(A).$$</span>
Thus, if the indifference curve is thick around point<span class="math-container">$A$</span> we must have <span class="math-container">$\lambda A+(1-\lambda) B\sim A\sim B$</span>. But this contradicts the strict convexity hypothesis. QED</p>

<hr>

<p>That should do the job for strict convexity. As frage_man already pointed out, the proposition fails to hold generally under weak convexity. Indifference everywhere is a very elegant example why.</p>

CC BY-SA 3.0
2016-08-21T11:58:14.407Z
I thought about a definition for thick IC like yours and the thing I was unsure about was whether we needed every point in the neighborhood to be indifferent, Because a point on the boundary of the thick IC is only indifferent to a `half-ball' in a sense. Thoughts?
VCG
CC BY-SA 3.0
2016-08-21T14:43:00.413Z
@frage_man Yes, this is why I don't require my definition to hold at every point on the IC. Instead, I require only that we can find *at least one* point in the interior of the IC's area around which we can draw a ball of indifference. As soon as that's true for one point, we know the IC isn't 'thin'. I can't think of a way to construct a thick IC that would have no such points at all. I edited the wording of my answer to be slightly clearer on this point.
Ubiquitous
CC BY-SA 3.0
2016-08-21T14:47:45.840Z
Gotcha - great answer btw.
VCG
CC BY-SA 4.0
2016-08-21T09:12:54.067Z
2018-11-19T07:25:17.407Z
2018-11-19T07:25:17.407Z
Ubiquitous
<p>It is standard in many micro textbooks when analyzing the relationship between preference axioms and the shape of the utility function (and consequently the shape of indifference curves), to attributed the "non-thickness" of indifference curves to "local non-satiation".
While it is easily seen (and proven) that LNS preferences don't admit thick IC, my question is the following:</p>

<p>Thick IC violate as well strict convexity (notice that weak convexity appears to "survive"), so, isn't convexity alone another property that prevents thickness? Put it differently, can we find non monotonic, convex preferences that don't admit thick IC? </p>

<p>If any proof or reference to relevant literature would be very useful.</p>

CC BY-SA 3.0
2016-08-20T22:03:11.553Z
Strict convexity at least implies that indifference curves have nonempty interior.
Michael Greinecker
<mathematical-economics><utility>
Doesnt convexity prevent thick indifference curves aswell?
thekiciminister
# Doesnt convexity prevent thick indifference curves aswell?

(Asked by: username_0 on 2016-08-20)

It is standard in many micro textbooks when analyzing the relationship between preference axioms and the shape of the utility function (and consequently the shape of indifference curves), to attributed the "non-thickness" of indifference curves to "local non-satiation".
While it is easily seen (and proven) that LNS preferences don't admit thick IC, my question is the following:

Thick IC violate as well strict convexity (notice that weak convexity appears to "survive"), so, isn't convexity alone another property that prevents thickness? Put it differently, can we find non monotonic, convex preferences that don't admit thick IC?

If any proof or reference to relevant literature would be very useful.

**Tags:** mathematical-economics, utility
> **Comments:**
> **username_1**: Strict convexity at least implies that indifference curves have nonempty interior.  

---

## 2 answers

### Answer by username_3 on 2016-08-21. Score: 8 | Accepted answer

My approach would be to define thick indifference curves in terms of a stronger form of local non-satiation:

> **Definition (thick indifference curves)** Preferences are said to have thick indifference curves if there exists at least one bundle $A\in\mathbb{R}^l$ and an open ball
> $\mathscr{B}(A)$ around $A$ such that $A'\sim A$ for *every*
> $A'\in\mathscr{B}(A)$.

Let's set up a definition for strict convexity to be sure we are on the same page:

> **Definition (strict convexity)** Preferences obey strict convexity if
>
> $$A\sim B\implies \lambda A+(1-\lambda) B\succ A\sim B$$
>
> for $A,B\in\mathbb{R}^l$ and any $\lambda\in(0,1)$.

Now we can state the desired result:

> **Proposition** If $\succsim$ is strictly convex then $\succsim$ does not have thick indifference curves.

**Proof** Suppose that $\succsim$ has thick indifference curves. Then there exists an $A$ such that $A'\sim A$ for all $A'\in\mathscr{B}(A)$. Thus, fix two bundles $A$ and $B$ such that $A\sim B$ and $B\in\mathscr{B}(A)$. We know that
$$\lambda A+(1-\lambda) B \in \mathscr{B}(A).$$
Thus, if the indifference curve is thick around point$A$ we must have $\lambda A+(1-\lambda) B\sim A\sim B$. But this contradicts the strict convexity hypothesis. QED

---

That should do the job for strict convexity. As frage\_man already pointed out, the proposition fails to hold generally under weak convexity. Indifference everywhere is a very elegant example why.

> **Comments:**
> **username_2**: I thought about a definition for thick IC like yours and the thing I was unsure about was whether we needed every point in the neighborhood to be indifferent, Because a point on the boundary of the thick IC is only indifferent to a `half-ball' in a sense. Thoughts?  
> **username_3**: @frage_man Yes, this is why I don't require my definition to hold at every point on the IC. Instead, I require only that we can find *at least one* point in the interior of the IC's area around which we can draw a ball of indifference. As soon as that's true for one point, we know the IC isn't 'thin'. I can't think of a way to construct a thick IC that would have no such points at all. I edited the wording of my answer to be slightly clearer on this point.  
> **username_2**: Gotcha - great answer btw.  

---

### Answer by username_2 on 2016-08-20. Score: 4

Great question, never thought about this:

So the main thing that LNS preferences get rid of is a satiation point, preventing the budget from binding in the consumer's problem. So there are tons of preferences that are non-monotonic, strictly convex, with a satiation point.

Now a simple example of weakly convex with thick IC is indfference everywhere. So $\forall x,y\in X, x\sim y$. In this case the IC is all of $\mathbb{R}$.

I agree with you on that strict convexity eliminates thick IC. I'm working on a proof but defining thick IC in mathmatical terms is tricky. However, just recall non-thick IC is not enough to guarantee LNS, so that's why strict convexity is not a substitute for LNS in major theorems like the 2nd Welfare Theorem.

> **Comments:**
> **username_0**: Thanks for the answer! Indifference everywhere is a great example of thick IC which satisfy weak convexity. I agree with you that while strong convexity may get rid of thick IC (proof pending, im trying to work out one myself too), they dont guarantee LNS, and the usefullness for some theorems vanishes. Nevertheless, I think that the proof would be of value by itslef, because it would asure that LNS is not the only property that can eliminate thickness, aspect that is not mentioned at all in micro textbooks, it just broadens our understanding of different axioms to the shape of u(.) functions  

---

