<p>Consider any utility function <span class="math-container">$u: \mathbb{R}^2_+ \rightarrow\mathbb{R}$</span>. Indifference curve for satisfaction level <span class="math-container">$\mu$</span> is defined as:
<span class="math-container">$\text{IC}(\mu) = \{(x, y)\in\mathbb{R}^2_+|u(x, y) = \mu\}$</span></p>
<p>I am assuming that by ICs cannot cross, you mean that ICs for two different utility levels don't have a non-empty intersection i.e.<br />
<span class="math-container">$(\forall (\mu_1, \mu_2)\in\mathbb{R}^2) (\mu_1 \neq \mu_2 \Rightarrow \text{IC}(\mu_1) \cap \text{IC}(\mu_2) = \emptyset)$</span>.</p>
<p>This is obviously true because of the fact that <span class="math-container">$u$</span> is a function.</p>
<p>Related Question: Can an IC look like a cross? <a href="https://qr.ae/pG0AoL" rel="nofollow noreferrer">https://qr.ae/pG0AoL</a></p>

CC BY-SA 4.0
2022-05-05T19:06:57.563Z
Regarding the link: if we assume either strong monotonicity or strict convexity then the example cannot be correct.
High GPA
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2022-05-06T02:09:38.243Z
Yes, that's right.
Amit
CC BY-SA 4.0
2022-05-06T11:21:35.643Z
In fact, AMit, the original motivation of this question is: "if an IC cannot be looked like a cross, then what are some assumptions behind that?". However, later I found that I had a misconception about ICs: I thought that, if the ICs look like a cross, then they can be different ICs. In fact, they can only be the same IC.
High GPA
CC BY-SA 4.0
2022-05-05T14:53:17.923Z
2022-05-05T14:53:17.923Z
Amit
<blockquote>
  <p>If only weak-ordering and continuity is assumed, ICs can definitely intersect.</p>
</blockquote>

<p>This is not true. First, if you're speaking of indifference <em>curves</em>, you'd already be assuming either local non-satiation or monotonicity. </p>

<p>Let's speak of indifference <em>sets</em> instead. The analogue of two sets, <span class="math-container">$I_1$</span> and <span class="math-container">$I_2$</span>, "crossing" each other can be formalized as <span class="math-container">$I_1\ne I_2$</span> and <span class="math-container">$I_1\cap I_2\ne \varnothing$</span>. </p>

<p>Take two alternatives <span class="math-container">$x_1,x_2\in X$</span> and define two indifference sets as follows:
<span class="math-container">\begin{align}
I_1:=\{x\in X:x\sim x_1\}\quad\text{and}\quad I_2:=\{x\in X:x\sim x_2\}. 
\end{align}</span>
WLOG, assume that <span class="math-container">$x_1\succ x_2$</span>, so that <span class="math-container">$I_1\ne I_2$</span>. If we allow <span class="math-container">$I_1$</span> to "cross" <span class="math-container">$I_2$</span>, then <span class="math-container">$I_1\cap I_2$</span> must be non-empty. Let <span class="math-container">$\bar x\in I_1\cap I_2$</span> be an element in the intersection. By the transitivity of the indifference relation <span class="math-container">$\sim$</span>, we have <span class="math-container">$x_1\sim \bar x$</span> and <span class="math-container">$\bar x\sim x_2$</span>, implying <span class="math-container">$x_1\sim x_2$</span>. But this is contradictory to our assumption that <span class="math-container">$x_1\succ x_2$</span>. The contradiction therefore suggests that transitivity, a property of the weak ordering, is violated. </p>

CC BY-SA 4.0
2019-03-07T23:15:09.197Z
Thanks for your answer! So what is the assumption needed for drawing indifference curves? Is weak-order + local non-satiation enough?
High GPA
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2019-03-07T23:21:25.943Z
@HighGPA: You're welcome. To have indifference *curves*, you'd need completeness, transitivity, continuity, and non-satiation. Non-satiation can be substituted with monotonicity if ICs are required to be downward sloping.
Herr K.
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2019-03-07T23:24:58.470Z
I guess your monotonicity here means strict monotonicity? Without non-satiation, Weak-order + Cont + Convexity or Weak-order + cont + Mono (not strict) seem to imply something like "indifference manifold"
High GPA
CC BY-SA 4.0
2019-03-07T23:31:22.390Z
@HighGPA: By monotonicity I mean if $x\ge x'$ (element by element), then $x\succsim x'$, and if $x>x'$ then $x\succ x'$.
Herr K.
CC BY-SA 4.0
2019-06-09T17:00:44.503Z
Hi Herr, I just learned how to prove that non-saitation (along with completeness, transitivity and continuity) implies that the indifference set must have empty interior. However I am still not sure how to show that the indifference set can only contain curves and cannot contain a cross
High GPA
CC BY-SA 4.0
2019-06-09T21:30:46.260Z
@HighGPA: If two indifference curves cross one another, that'll produce a violation of transitivity, which is what I showed in my answer above. Is there a specific step that you don't follow? Or perhaps I misunderstood your question?
Herr K.
CC BY-SA 4.0
2019-06-10T09:12:12.983Z
So continuity, completeness, and transitivity give indifference set, not curves. For example the utility function $u(x,y)=C$ where $C$ is constant. How do we move from "set" to "curve" by assuming non-satiation?
High GPA
CC BY-SA 4.0
2019-06-10T15:25:50.170Z
@HighGPA: A constant utility function violates non satiation. But if you assume $u(x,y)=x +y$ (or other common utility functions that satisfy non satiation), then you get indifference curves.
Herr K.
CC BY-SA 4.0
2019-03-07T23:02:09.543Z
2019-03-07T23:02:09.543Z
Herr K.
<p>If <strong>only</strong> weak-ordering and continuity is assumed, "ICs" can definitely intersect.</p>

<p>If we assume Monotonicity or convexity in addition to weak-ordering, then we can get "no cross of IC".</p>

<p>But those two assumptions are argueably too strong. In some real world scenarios they are violated. </p>

<p>Are there any weaker assumptions for the "IC cannot cross"?  </p>

CC BY-SA 4.0
2019-03-07T22:52:24.753Z
1. Convexity is no problem. In case of $U(x,y) = 0$ the IC is the whole plane. This is a convex preference relation, and no "ICs cross" as there is only one IC.
Giskard
CC BY-SA 4.0
2019-03-07T22:52:48.373Z
2. The problem with your question is that "weaker" is not a mathematical concept in this context. Would you accept transitivity as a weaker condition? Please edit your question to clarify.
Giskard
<microeconomics><utility>
What is the assumption behind "indifference curve does not cross"?
High GPA
# What is the assumption behind "indifference curve does not cross"?

(Asked by: username_0 on 2019-03-07)

If **only** weak-ordering and continuity is assumed, "ICs" can definitely intersect.

If we assume Monotonicity or convexity in addition to weak-ordering, then we can get "no cross of IC".

But those two assumptions are argueably too strong. In some real world scenarios they are violated.

Are there any weaker assumptions for the "IC cannot cross"?

**Tags:** microeconomics, utility

> **Comments:**
> **username_1**: 1. Convexity is no problem. In case of $U(x,y) = 0$ the IC is the whole plane. This is a convex preference relation, and no "ICs cross" as there is only one IC.  
> **username_1**: 2. The problem with your question is that "weaker" is not a mathematical concept in this context. Would you accept transitivity as a weaker condition? Please edit your question to clarify.  

---

## 2 answers

### Answer by username_3 on 2019-03-07. Score: 6 | Accepted answer

> If only weak-ordering and continuity is assumed, ICs can definitely intersect.

This is not true. First, if you're speaking of indifference *curves*, you'd already be assuming either local non-satiation or monotonicity.

Let's speak of indifference *sets* instead. The analogue of two sets, $I\_1$ and $I\_2$, "crossing" each other can be formalized as $I\_1\ne I\_2$ and $I\_1\cap I\_2\ne \varnothing$.

Take two alternatives $x\_1,x\_2\in X$ and define two indifference sets as follows:
\begin{align}
I\_1:=\{x\in X:x\sim x\_1\}\quad\text{and}\quad I\_2:=\{x\in X:x\sim x\_2\}.
\end{align}
WLOG, assume that $x\_1\succ x\_2$, so that $I\_1\ne I\_2$. If we allow $I\_1$ to "cross" $I\_2$, then $I\_1\cap I\_2$ must be non-empty. Let $\bar x\in I\_1\cap I\_2$ be an element in the intersection. By the transitivity of the indifference relation $\sim$, we have $x\_1\sim \bar x$ and $\bar x\sim x\_2$, implying $x\_1\sim x\_2$. But this is contradictory to our assumption that $x\_1\succ x\_2$. The contradiction therefore suggests that transitivity, a property of the weak ordering, is violated.

> **Comments:**
> **username_0**: Thanks for your answer! So what is the assumption needed for drawing indifference curves? Is weak-order + local non-satiation enough?  
> **username_3**: @HighGPA: You're welcome. To have indifference *curves*, you'd need completeness, transitivity, continuity, and non-satiation. Non-satiation can be substituted with monotonicity if ICs are required to be downward sloping.  
> **username_0**: I guess your monotonicity here means strict monotonicity? Without non-satiation, Weak-order + Cont + Convexity or Weak-order + cont + Mono (not strict) seem to imply something like "indifference manifold"  
> **username_3**: @HighGPA: By monotonicity I mean if $x\ge x'$ (element by element), then $x\succsim x'$, and if $x>x'$ then $x\succ x'$.  
> **username_0**: Hi Herr, I just learned how to prove that non-saitation (along with completeness, transitivity and continuity) implies that the indifference set must have empty interior. However I am still not sure how to show that the indifference set can only contain curves and cannot contain a cross  
> **username_3**: @HighGPA: If two indifference curves cross one another, that'll produce a violation of transitivity, which is what I showed in my answer above. Is there a specific step that you don't follow? Or perhaps I misunderstood your question?  
> **username_0**: So continuity, completeness, and transitivity give indifference set, not curves. For example the utility function $u(x,y)=C$ where $C$ is constant. How do we move from "set" to "curve" by assuming non-satiation?  
> **username_3**: @HighGPA: A constant utility function violates non satiation. But if you assume $u(x,y)=x +y$ (or other common utility functions that satisfy non satiation), then you get indifference curves.  

---

### Answer by username_2 on 2022-05-05. Score: 2

Consider any utility function $u: \mathbb{R}^2\_+ \rightarrow\mathbb{R}$. Indifference curve for satisfaction level $\mu$ is defined as:
$\text{IC}(\mu) = \{(x, y)\in\mathbb{R}^2\_+|u(x, y) = \mu\}$

I am assuming that by ICs cannot cross, you mean that ICs for two different utility levels don't have a non-empty intersection i.e.  
$(\forall (\mu\_1, \mu\_2)\in\mathbb{R}^2) (\mu\_1 \neq \mu\_2 \Rightarrow \text{IC}(\mu\_1) \cap \text{IC}(\mu\_2) = \emptyset)$.

This is obviously true because of the fact that $u$ is a function.

Related Question: Can an IC look like a cross? <https://qr.ae/pG0AoL>

> **Comments:**
> **username_0**: Regarding the link: if we assume either strong monotonicity or strict convexity then the example cannot be correct.  
> **username_2**: Yes, that's right.  
> **username_0**: In fact, AMit, the original motivation of this question is: "if an IC cannot be looked like a cross, then what are some assumptions behind that?". However, later I found that I had a misconception about ICs: I thought that, if the ICs look like a cross, then they can be different ICs. In fact, they can only be the same IC.  

---

