<p>To begin with, I think the question is wrongly stated. For if the defininition of a thin indifference curve is such that continuity of a consumer's preferences implies thin indifference curves, then, surely, continuity implies thin indifference curves... This answers your question.</p>

<p>However, if we are to make a suitable definition of a thin indifference curve, we may firstly say that $$[q]=\{p\in\Delta|p\sim q\}$$ is a <em>thick</em> indifference curve, where $\Delta$ is the set of possible bundles, and where $\sim$ denotes indifference, whenever there exists a $q'\in [q]$ and a $\epsilon&gt;0$ such that $p\in N_{\epsilon}(q')$ implies $p\sim q'$, where $N_{\epsilon}(q')$ is some epsilon-neighbourhood around $q'$; and secondly say that $[q]$ is a <em>thin</em> indifference curve if it is not thick. Colloquially this means that there is some bump on a thick indifference curve $[q]$, but no such bump on a thin indifference curve.</p>

<p>Essentially, the above is a short exposition of <a href="http://www.econ.la.psu.edu/people-documents/k1c/Kalyan%20Chatterjee%20-%20A%20Geometric%20Approach%20to%20Expected%20Utility.pdf" rel="nofollow noreferrer">A Geometric Approach to Expected Utility (Chatterjee &amp; Krishna, 2006)</a>. Using the above definition of a thin indifference curve, they show in Lemma 2.3 that (i) continuity and (ii) independence implies thin indifference curves (note that they do not show that continuity alone implies thin indifference curves; cf. Ubiquitous' answer). Their definition relies on the following two topological concepts. </p>

<ol>
<li><em>The assumption of continuity.</em> All subsets $\{q|q\succ p\}$ and $\{q|p\succ q\}$ of $\Delta$, where $p\in \Delta$, are open; here, remember that an open set is a set for which each point in it has a neighbourhood lying in that set. Thus, this notion of continuity is similar to yours. </li>
<li><em>The assumption of independence.</em> For all $p,q,r\in\Delta$, $p\succ q$ and $\lambda\in (0,1]$ implies that
$$\lambda p+(1-\lambda)r\succ \lambda
q+(1-\lambda)r;$$ this allows for some nice algebra.</li>
</ol>

<p>Now, what they show in Lemma 2.3 is essentially that if you have an indifference curve $[q]$ and consider some epsilon-neighbourhood $N_{\epsilon}(q')$ around $q'\in [q]$, then $p\in N_{\epsilon}(q')$ will not imply that $p\sim q'$ for arbitrarily small $\epsilon&gt;0$. I.e., however small, no epsilon-neighbourhood is such that it only contains bundles for which one is indifferent between those bundles and $q'$. Instead, every epsilon-neighborhood will include points that are strictly preferred to $q'$.</p>

<p>For continuous utility functions, I think it is fruitful to note that their image in e.g. $\mathbb{R}^2$ has (Lebesgue) measure 0 (cf. <a href="https://math.stackexchange.com/questions/221110/how-to-prove-that-the-image-of-a-continuous-curve-in-mathbbr2-has-measure">How to prove that the image of a continuous curve in $\mathbb{R}^2$ has measure $0$?</a>)</p>

CC BY-SA 3.0
2016-12-02T09:29:34.677Z
2016-12-03T14:37:48.597Z
2017-04-13T12:19:37.213Z
Elias
<p>I don't think continuity alone is enough to guarantee thin indifference curves.</p>

<p>Consider preferences such that, for any $x$ and $y$ in the choice set, the consumer is indifferent between $x$ and $y$. This seems like it must fit any definition of a thick indifference curve because the whole choice set lies on a single indifference curve!</p>

<p>But these preferences also satisfy your definition of continuity.</p>

<p>Thus, it seems like continuity only implies thin indifference curves if it is paired with some other assumption.</p>

CC BY-SA 3.0
2016-12-02T18:40:44.583Z
2016-12-02T18:40:44.583Z
Ubiquitous
<p>If a consumer follows the rationality axiom of continuity (i.e. no jumps in his preferences), the indifference curves of a utility function are said to be thin.</p>

<p>Why does continuity ($x \succeq y \Rightarrow \exists \space z=x+\epsilon$ such that $|z|\ge y \space \forall \epsilon &gt; 0$) imply thin indifference curves?</p>

CC BY-SA 3.0
2016-11-02T18:04:17.837Z
Try these lecture notes: http://www.econ.ucla.edu/sboard/teaching/econ11_09/econ11_09_lecture2.pdf
usul
<microeconomics><utility><consumer-theory><preferences>
Thin indifference curves
Omrane
# Thin indifference curves

(Asked by: username_0 on 2016-10-30)

If a consumer follows the rationality axiom of continuity (i.e. no jumps in his preferences), the indifference curves of a utility function are said to be thin.

Why does continuity ($x \succeq y \Rightarrow \exists \space z=x+\epsilon$ such that $|z|\ge y \space \forall \epsilon > 0$) imply thin indifference curves?

**Tags:** microeconomics, utility, consumer-theory, preferences
> **Comments:**
> **username_1**: Try these lecture notes: http://www.econ.ucla.edu/sboard/teaching/econ11_09/econ11_09_lecture2.pdf  

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## 2 answers

### Answer by username_3 on 2016-12-02. Score: 6 | Accepted answer

I don't think continuity alone is enough to guarantee thin indifference curves.

Consider preferences such that, for any $x$ and $y$ in the choice set, the consumer is indifferent between $x$ and $y$. This seems like it must fit any definition of a thick indifference curve because the whole choice set lies on a single indifference curve!

But these preferences also satisfy your definition of continuity.

Thus, it seems like continuity only implies thin indifference curves if it is paired with some other assumption.


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### Answer by username_2 on 2016-12-02. Score: 6

To begin with, I think the question is wrongly stated. For if the defininition of a thin indifference curve is such that continuity of a consumer's preferences implies thin indifference curves, then, surely, continuity implies thin indifference curves... This answers your question.

However, if we are to make a suitable definition of a thin indifference curve, we may firstly say that $$[q]=\{p\in\Delta|p\sim q\}$$ is a *thick* indifference curve, where $\Delta$ is the set of possible bundles, and where $\sim$ denotes indifference, whenever there exists a $q'\in [q]$ and a $\epsilon>0$ such that $p\in N\_{\epsilon}(q')$ implies $p\sim q'$, where $N\_{\epsilon}(q')$ is some epsilon-neighbourhood around $q'$; and secondly say that $[q]$ is a *thin* indifference curve if it is not thick. Colloquially this means that there is some bump on a thick indifference curve $[q]$, but no such bump on a thin indifference curve.

Essentially, the above is a short exposition of [A Geometric Approach to Expected Utility (Chatterjee & Krishna, 2006)](http://www.econ.la.psu.edu/people-documents/k1c/Kalyan%20Chatterjee%20-%20A%20Geometric%20Approach%20to%20Expected%20Utility.pdf). Using the above definition of a thin indifference curve, they show in Lemma 2.3 that (i) continuity and (ii) independence implies thin indifference curves (note that they do not show that continuity alone implies thin indifference curves; cf. Ubiquitous' answer). Their definition relies on the following two topological concepts.

1. *The assumption of continuity.* All subsets $\{q|q\succ p\}$ and $\{q|p\succ q\}$ of $\Delta$, where $p\in \Delta$, are open; here, remember that an open set is a set for which each point in it has a neighbourhood lying in that set. Thus, this notion of continuity is similar to yours.
2. *The assumption of independence.* For all $p,q,r\in\Delta$, $p\succ q$ and $\lambda\in (0,1]$ implies that
   $$\lambda p+(1-\lambda)r\succ \lambda
   q+(1-\lambda)r;$$ this allows for some nice algebra.

Now, what they show in Lemma 2.3 is essentially that if you have an indifference curve $[q]$ and consider some epsilon-neighbourhood $N\_{\epsilon}(q')$ around $q'\in [q]$, then $p\in N\_{\epsilon}(q')$ will not imply that $p\sim q'$ for arbitrarily small $\epsilon>0$. I.e., however small, no epsilon-neighbourhood is such that it only contains bundles for which one is indifferent between those bundles and $q'$. Instead, every epsilon-neighborhood will include points that are strictly preferred to $q'$.

For continuous utility functions, I think it is fruitful to note that their image in e.g. $\mathbb{R}^2$ has (Lebesgue) measure 0 (cf. [How to prove that the image of a continuous curve in $\mathbb{R}^2$ has measure $0$?](https://math.stackexchange.com/questions/221110/how-to-prove-that-the-image-of-a-continuous-curve-in-mathbbr2-has-measure))


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