<p>Given your last comment above it seems that what you are really asking is whether the indifference sets of a continuous preference relation on <span class="math-container">$\mathbb R^n_+$</span> are <a href="https://en.wikipedia.org/wiki/Connected_space#Path_connectedness" rel="noreferrer">path-connected</a>. The answer is <em>No</em>. Let <span class="math-container">$n=1$</span> and let the preference relation be represented by <span class="math-container">$u(x)=(1-x)^2$</span>. Then the indifference set e.g. for <span class="math-container">$u=1$</span> is <span class="math-container">$\{0\}\cup\{2\}$</span>, which is not path-connected.</p>

CC BY-SA 4.0
2021-07-09T02:03:36.263Z
Thanks for the example! I assumed monotonicity (non-satiation) and was wondering how could a curve be not path-connected when an IC is continuous (and non-"thick"). With these assumptions, I chose the identity function without realizing the potential issue which is the domain is then only a subset of $X$ that is $I$. (This takes into account the case when $|I| < \infty$ as in your example.)
not tdm's twin
CC BY-SA 4.0
2021-07-09T02:05:38.077Z
**Def.** *Thick* Preferences are said to have *thick* ICs if $\exists$ at least one bundle $A \in X$ and an open ball $B(A)$ around $A$ such that $A′ \sim A$ for every $A′ \in B(A)$. [[Courtesy](https://economics.stackexchange.com/a/13134/33107)]
not tdm's twin
CC BY-SA 4.0
2021-07-09T02:07:32.727Z
I am sorry although your answer was extremely helpful, I will have to accept the other answer as it clarifies a few more questions I had, including this.
not tdm's twin
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2021-07-09T07:35:05.470Z
@dictatemetokcus, no problem!
VARulle
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2021-07-08T13:10:34.930Z
2021-07-08T13:10:34.930Z
VARulle
<p>To follow up on the answer of @VARulle let me give you some conditions for which the indifference curve is path connected.</p>
<p>The argument can also be found in the book <a href="https://rads.stackoverflow.com/amzn/click/com/0521585295" rel="noreferrer" rel="nofollow noreferrer">Mathematical Methods and Models for Economists</a> by Angel de la Fuente.</p>
<p>Preferences are <strong>monotone</strong> if <span class="math-container">$x &gt; y$</span> implies <span class="math-container">$x \succ y$</span> and that preferences are <strong>continuous</strong> if <span class="math-container">$x_n \succeq y_n$</span>, <span class="math-container">$x_n \to x$</span> and <span class="math-container">$y_n \to y$</span> imply <span class="math-container">$x \succeq y$</span>.</p>
<p>Let us limit ourselves to the space <span class="math-container">$\mathbb{R}^n_{++}$</span> of strictly positive bundles. For some <span class="math-container">$a \in \mathbb{R}^n_{++}$</span> consider the indifference curve <span class="math-container">$I(a) = \{x \in \mathbb{R}^n_{++}| x \sim a\}$</span>.</p>
<p><strong>Theorem</strong> If preferences are monotone and continuous then for <span class="math-container">$a \in \mathbb{R}^n_{++}$</span>,  the set <span class="math-container">$I(a)$</span> is path-connected.</p>
<p>As being path-connected is a topological property, it suffices to show that there is a homeomorphism between <span class="math-container">$I(a)$</span> and a path-connected set. For this, we will take the set <span class="math-container">$\Delta$</span>:
<span class="math-container">$$
\Delta = \{z \in \mathbb{R}^n_{++}: \sum_i z_i = 1\}
$$</span>
This set is path-connected as any two elements in <span class="math-container">$\Delta$</span> can be connected by the line segment that has these points as endpoints.</p>
<p>Consider the radial projection function <span class="math-container">$f: I(a) \to \Delta$</span> defined as:
<span class="math-container">$$
f(x) = \dfrac{x}{\sum_i x_i}
$$</span>
As we restrict ourselves to <span class="math-container">$\mathbb{R}^n_{++}$</span>, the function <span class="math-container">$f$</span> is well defined. It is easy to see that <span class="math-container">$f$</span> is continuous. To show that it is a homeomorphism, we have to show that it is a bijection and that <span class="math-container">$f^{-1}$</span> is also continuous.</p>
<p>First, to show that <span class="math-container">$f$</span> is onto. let <span class="math-container">$z \in \Delta$</span> and consider the ray <span class="math-container">$\alpha z$</span> with <span class="math-container">$\alpha &gt; 0$</span>. Then for <span class="math-container">$\alpha$</span> large enough, we have <span class="math-container">$\alpha z &gt; a$</span>, so by monotonicity <span class="math-container">$\alpha z \succ a$</span>. Also for <span class="math-container">$\alpha &gt; 0$</span> small enough, we have <span class="math-container">$\alpha z &lt; a$</span> so, by monotonicity, <span class="math-container">$a \succ \alpha z$</span>.</p>
<p>Let <span class="math-container">$\alpha^\ast = \inf \{\alpha| \alpha z \succ a\}$</span> which is well defined by the argument above. Then by continuity <span class="math-container">$\alpha^\ast z \succeq a$</span>. Also for all <span class="math-container">$\alpha &lt; \alpha^\ast$</span> we have <span class="math-container">$a  \succeq \alpha z$</span>  so again by continuity <span class="math-container">$a \succeq \alpha^\ast z$</span>. This shows that <span class="math-container">$\alpha^\ast z \sim a$</span> or equivalently <span class="math-container">$\alpha^\ast z \in I(a)$</span>. Defining <span class="math-container">$x = \alpha^\ast z$</span>, gives that <span class="math-container">$x \sim a$</span> and <span class="math-container">$h(x) = z$</span>, so <span class="math-container">$h$</span> is onto.</p>
<p>To show that <span class="math-container">$f$</span> is a bijection, let <span class="math-container">$f(x) = f(y) = z$</span>. We need to show that <span class="math-container">$x = y$</span>. We have:
<span class="math-container">$$
\dfrac{x}{\sum_i x_i} = \dfrac{y}{\sum_i y_i} = z.
$$</span>
As such,
<span class="math-container">$$
x \sum_i y_i = y \sum_i x_i,
$$</span>
which shows that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are proportional to each other, i.e.,
<span class="math-container">$$
\alpha x = \beta y.
$$</span>
if <span class="math-container">$\alpha = \beta$</span> then <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are identical, so either <span class="math-container">$\alpha &gt; \beta$</span> or vice versa. But then by monotonicity <span class="math-container">$y \succ  \frac{\beta}{\alpha} y = x$</span> or vice versa, which gives a contradiction with <span class="math-container">$x, y \in I(a)$</span>.</p>
<p>Finally, to see that <span class="math-container">$f^{-1}$</span> is continuous, let <span class="math-container">$z \in \Delta$</span> and take any sequence <span class="math-container">$z_n \to z$</span>. It suffices to show the existence of a subsequence <span class="math-container">$z_{k_n}$</span> such that <span class="math-container">$h^{-1}(z_{k_n}) \to h^{-1}(z)$</span>.</p>
<p>Define <span class="math-container">$x_n = h^{-1}(z_n)$</span> and <span class="math-container">$x = h^{-1}(z)$</span>. Then:
<span class="math-container">$$
\alpha_n x_n = z_n.
$$</span>
with <span class="math-container">$\alpha_n = \frac{1}{\sum_i x_{i,n}}$</span> also we have that:
<span class="math-container">$$
\alpha x = z,
$$</span>
with <span class="math-container">$\alpha = \frac{1}{\sum_i x_i}$</span>. It suffices to show that there is a subsequence for which <span class="math-container">$x_{k_n} \to x$</span>.</p>
<p>If <span class="math-container">$\{\alpha_n\}$</span> is bounded, there is a convergent subsequence <span class="math-container">$\alpha_{k_n} \to \alpha^\ast$</span>. If <span class="math-container">$\alpha^\ast = 0$</span> then as <span class="math-container">$z_n \to z$</span> there is an <span class="math-container">$n_k$</span> large enough such that <span class="math-container">$x_{n_k} = \frac{z_{n_k}}{\alpha_{n_k}} &gt; \frac{z}{\alpha} = x$</span> which would contradict <span class="math-container">$x_{n_k} \sim x$</span>. As such, <span class="math-container">$\alpha^\ast &gt; 0$</span> and:
<span class="math-container">$$
x_{k_n} = \frac{z_{k_n}}{\alpha_{k_n}} \to \frac{z}{\alpha^\ast} = \frac{\alpha}{\alpha^\ast} x.
$$</span>
Then as <span class="math-container">$x_{k_n} \sim x$</span> for all <span class="math-container">$n$</span>, by continuity, <span class="math-container">$x \sim \frac{\alpha}{\alpha^\ast}x$</span> which can only hold (by monotonicity) if <span class="math-container">$\alpha^\ast = \alpha$</span>. This shows that  <span class="math-container">$x_n \to x$</span>.</p>
<p>If <span class="math-container">$\alpha_n &gt; 0$</span> is unbounded, then there is a subsequence <span class="math-container">$\alpha_{k_n} \to \infty$</span>. Then along this subsequence (as <span class="math-container">$z_{k_n} \to z$</span>) we have
<span class="math-container">$$
x_{k_n} = \frac{z_{k_n}}{\alpha_{k_n}} \to 0,
$$</span>
So for <span class="math-container">$n$</span> large enough <span class="math-container">$x_{k_n} &lt; x$</span> which means that <span class="math-container">$x \succ x_{k_n}$</span> which again contradicts <span class="math-container">$x_{k_n} \sim x$</span>. This shows that <span class="math-container">$f^{-1}$</span> is continuous <span class="math-container">$\square$</span>.</p>

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2021-07-09T01:58:29.093Z
Ah, that's a very well-explained proof. I assumed monotonicity (non-satiation), and was wondering how could a curve be not path-connected when an IC is continuous (and non-"thick"). With these assumptions, I chose the function without realizing the potential issue which is the domain is then only a subset of $X$ that is $I$.
not tdm's twin
CC BY-SA 4.0
2021-07-09T02:00:59.193Z
This^ takes into account the case when $|I| < \infty$, as in VARulle's example. I will accept this answer once I read the complete argument, thank you for the explanation.
not tdm's twin
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2023-05-25T04:06:33.717Z
The theorem states that continuity of preference implies path-connectivity. I wonder if the converse is also true.
High GPA
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2021-07-08T14:06:39.507Z
2021-07-08T14:26:17.040Z
2021-07-08T14:26:17.040Z
tdm
<p>Assume the relation <span class="math-container">$\succeq$</span> is continuous (by the preferential definition). Does this mean the graph of Indifference Curves are continuous?</p>
<p>Since <span class="math-container">$\sim$</span> satisfies the definition for <span class="math-container">$\succeq$</span>, we have the IC follow the preferential definition. But what about the topological definition?</p>
<p>I think it does, and here's what I have done. Please verify!</p>
<blockquote>
<p>Rephrasing the question: Given <span class="math-container">$I = \{x : x \sim y\}$</span> equipped with the standard topology, consider <span class="math-container">$f : I \to \mathbb{R}^n$</span> such that <span class="math-container">$f(x) = x$</span>. How do you show <span class="math-container">$f$</span> is continuous? (<strong>Note.</strong> I don't know if it is true, I have assumed it to be true and tried to show that it indeed is.)</p>
<p><em>Proof.</em> An arbitrary <span class="math-container">$S \subseteq I$</span> when closed does not violate the continuity, so for any other <span class="math-container">$S$</span> which is open, <span class="math-container">$f^{-1}(S) = S$</span> is open. <span class="math-container">$\blacksquare$</span>.</p>
</blockquote>

CC BY-SA 4.0
2021-07-08T10:01:01.437Z
What exactly do you mean with a "continuous graph"? The $f$ you define is the identity function on $I$. This function is trivially continuous but has nothing to do with ICs.
VARulle
CC BY-SA 4.0
2021-07-08T10:38:46.037Z
@VARulle Well, so if you plot a particular IC, say IC(y) = {x : x ~ y}, then the function IC(y) is continuous. Note that I am considering the function IC and not a binary relation. I know the identity function is immediate, but isn't it the plot of IC the identity function I wrote?
not tdm's twin
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2021-07-08T11:15:11.610Z
@dictatemetokcus as you define it, $IC(y)$ produces a set (the set of bundles indifferent to $y$) and not a single bundle. This means that $IC(.)$ is a **correspondence** and not a function. You want to show that $IC$ is upper and lower hemi-continuous?
tdm
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2021-07-08T11:55:33.260Z
@tdm Sorry, that's my fault. So I was essentially trying to find out if while plotting the Indifference Curve, I can draw it without having to lift my pen(cil). While it is a set, we can represent the "plot" by defining f : {x : x ~ y} to R$^n$ given by f(x) = x. Isn't it so? Because, if we can represent it via the function f, we can prove (as I did in the post) that we can draw the curve without having to lift my pencil. I hope I was able to frame my question better this time.
not tdm's twin
<microeconomics><preferences>
Are Indifference Curve graphs continuous given the preferential definition of continuity?
not tdm's twin
# Are Indifference Curve graphs continuous given the preferential definition of continuity?

(Asked by: username_0 on 2021-07-08)

Assume the relation $\succeq$ is continuous (by the preferential definition). Does this mean the graph of Indifference Curves are continuous?

Since $\sim$ satisfies the definition for $\succeq$, we have the IC follow the preferential definition. But what about the topological definition?

I think it does, and here's what I have done. Please verify!

> Rephrasing the question: Given $I = \{x : x \sim y\}$ equipped with the standard topology, consider $f : I \to \mathbb{R}^n$ such that $f(x) = x$. How do you show $f$ is continuous? (**Note.** I don't know if it is true, I have assumed it to be true and tried to show that it indeed is.)
>
> *Proof.* An arbitrary $S \subseteq I$ when closed does not violate the continuity, so for any other $S$ which is open, $f^{-1}(S) = S$ is open. $\blacksquare$.

**Tags:** microeconomics, preferences
> **Comments:**
> **username_1**: What exactly do you mean with a "continuous graph"? The $f$ you define is the identity function on $I$. This function is trivially continuous but has nothing to do with ICs.  
> **username_0**: @VARulle Well, so if you plot a particular IC, say IC(y) = {x : x ~ y}, then the function IC(y) is continuous. Note that I am considering the function IC and not a binary relation. I know the identity function is immediate, but isn't it the plot of IC the identity function I wrote?  
> **username_2**: @dictatemetokcus as you define it, $IC(y)$ produces a set (the set of bundles indifferent to $y$) and not a single bundle. This means that $IC(.)$ is a **correspondence** and not a function. You want to show that $IC$ is upper and lower hemi-continuous?  
> **username_0**: @tdm Sorry, that's my fault. So I was essentially trying to find out if while plotting the Indifference Curve, I can draw it without having to lift my pen(cil). While it is a set, we can represent the "plot" by defining f : {x : x ~ y} to R$^n$ given by f(x) = x. Isn't it so? Because, if we can represent it via the function f, we can prove (as I did in the post) that we can draw the curve without having to lift my pencil. I hope I was able to frame my question better this time.  

---

## 2 answers

### Answer by username_2 on 2021-07-08. Score: 7 | Accepted answer

To follow up on the answer of @VARulle let me give you some conditions for which the indifference curve is path connected.

The argument can also be found in the book [Mathematical Methods and Models for Economists](https://rads.stackoverflow.com/amzn/click/com/0521585295) by Angel de la Fuente.

Preferences are **monotone** if $x > y$ implies $x \succ y$ and that preferences are **continuous** if $x\_n \succeq y\_n$, $x\_n \to x$ and $y\_n \to y$ imply $x \succeq y$.

Let us limit ourselves to the space $\mathbb{R}^n\_{++}$ of strictly positive bundles. For some $a \in \mathbb{R}^n\_{++}$ consider the indifference curve $I(a) = \{x \in \mathbb{R}^n\_{++}| x \sim a\}$.

**Theorem** If preferences are monotone and continuous then for $a \in \mathbb{R}^n\_{++}$, the set $I(a)$ is path-connected.

As being path-connected is a topological property, it suffices to show that there is a homeomorphism between $I(a)$ and a path-connected set. For this, we will take the set $\Delta$:
$$
\Delta = \{z \in \mathbb{R}^n\_{++}: \sum\_i z\_i = 1\}
$$
This set is path-connected as any two elements in $\Delta$ can be connected by the line segment that has these points as endpoints.

Consider the radial projection function $f: I(a) \to \Delta$ defined as:
$$
f(x) = \dfrac{x}{\sum\_i x\_i}
$$
As we restrict ourselves to $\mathbb{R}^n\_{++}$, the function $f$ is well defined. It is easy to see that $f$ is continuous. To show that it is a homeomorphism, we have to show that it is a bijection and that $f^{-1}$ is also continuous.

First, to show that $f$ is onto. let $z \in \Delta$ and consider the ray $\alpha z$ with $\alpha > 0$. Then for $\alpha$ large enough, we have $\alpha z > a$, so by monotonicity $\alpha z \succ a$. Also for $\alpha > 0$ small enough, we have $\alpha z < a$ so, by monotonicity, $a \succ \alpha z$.

Let $\alpha^\ast = \inf \{\alpha| \alpha z \succ a\}$ which is well defined by the argument above. Then by continuity $\alpha^\ast z \succeq a$. Also for all $\alpha < \alpha^\ast$ we have $a \succeq \alpha z$ so again by continuity $a \succeq \alpha^\ast z$. This shows that $\alpha^\ast z \sim a$ or equivalently $\alpha^\ast z \in I(a)$. Defining $x = \alpha^\ast z$, gives that $x \sim a$ and $h(x) = z$, so $h$ is onto.

To show that $f$ is a bijection, let $f(x) = f(y) = z$. We need to show that $x = y$. We have:
$$
\dfrac{x}{\sum\_i x\_i} = \dfrac{y}{\sum\_i y\_i} = z.
$$
As such,
$$
x \sum\_i y\_i = y \sum\_i x\_i,
$$
which shows that $x$ and $y$ are proportional to each other, i.e.,
$$
\alpha x = \beta y.
$$
if $\alpha = \beta$ then $x$ and $y$ are identical, so either $\alpha > \beta$ or vice versa. But then by monotonicity $y \succ \frac{\beta}{\alpha} y = x$ or vice versa, which gives a contradiction with $x, y \in I(a)$.

Finally, to see that $f^{-1}$ is continuous, let $z \in \Delta$ and take any sequence $z\_n \to z$. It suffices to show the existence of a subsequence $z\_{k\_n}$ such that $h^{-1}(z\_{k\_n}) \to h^{-1}(z)$.

Define $x\_n = h^{-1}(z\_n)$ and $x = h^{-1}(z)$. Then:
$$
\alpha\_n x\_n = z\_n.
$$
with $\alpha\_n = \frac{1}{\sum\_i x\_{i,n}}$ also we have that:
$$
\alpha x = z,
$$
with $\alpha = \frac{1}{\sum\_i x\_i}$. It suffices to show that there is a subsequence for which $x\_{k\_n} \to x$.

If $\{\alpha\_n\}$ is bounded, there is a convergent subsequence $\alpha\_{k\_n} \to \alpha^\ast$. If $\alpha^\ast = 0$ then as $z\_n \to z$ there is an $n\_k$ large enough such that $x\_{n\_k} = \frac{z\_{n\_k}}{\alpha\_{n\_k}} > \frac{z}{\alpha} = x$ which would contradict $x\_{n\_k} \sim x$. As such, $\alpha^\ast > 0$ and:
$$
x\_{k\_n} = \frac{z\_{k\_n}}{\alpha\_{k\_n}} \to \frac{z}{\alpha^\ast} = \frac{\alpha}{\alpha^\ast} x.
$$
Then as $x\_{k\_n} \sim x$ for all $n$, by continuity, $x \sim \frac{\alpha}{\alpha^\ast}x$ which can only hold (by monotonicity) if $\alpha^\ast = \alpha$. This shows that $x\_n \to x$.

If $\alpha\_n > 0$ is unbounded, then there is a subsequence $\alpha\_{k\_n} \to \infty$. Then along this subsequence (as $z\_{k\_n} \to z$) we have
$$
x\_{k\_n} = \frac{z\_{k\_n}}{\alpha\_{k\_n}} \to 0,
$$
So for $n$ large enough $x\_{k\_n} < x$ which means that $x \succ x\_{k\_n}$ which again contradicts $x\_{k\_n} \sim x$. This shows that $f^{-1}$ is continuous $\square$.

> **Comments:**
> **username_0**: Ah, that's a very well-explained proof. I assumed monotonicity (non-satiation), and was wondering how could a curve be not path-connected when an IC is continuous (and non-"thick"). With these assumptions, I chose the function without realizing the potential issue which is the domain is then only a subset of $X$ that is $I$.  
> **username_0**: This^ takes into account the case when $|I| < \infty$, as in VARulle's example. I will accept this answer once I read the complete argument, thank you for the explanation.  
> **username_3**: The theorem states that continuity of preference implies path-connectivity. I wonder if the converse is also true.  

---

### Answer by username_1 on 2021-07-08. Score: 7

Given your last comment above it seems that what you are really asking is whether the indifference sets of a continuous preference relation on $\mathbb R^n\_+$ are [path-connected](https://en.wikipedia.org/wiki/Connected_space#Path_connectedness). The answer is *No*. Let $n=1$ and let the preference relation be represented by $u(x)=(1-x)^2$. Then the indifference set e.g. for $u=1$ is $\{0\}\cup\{2\}$, which is not path-connected.

> **Comments:**
> **username_0**: Thanks for the example! I assumed monotonicity (non-satiation) and was wondering how could a curve be not path-connected when an IC is continuous (and non-"thick"). With these assumptions, I chose the identity function without realizing the potential issue which is the domain is then only a subset of $X$ that is $I$. (This takes into account the case when $|I| < \infty$ as in your example.)  
> **username_0**: **Def.** *Thick* Preferences are said to have *thick* ICs if $\exists$ at least one bundle $A \in X$ and an open ball $B(A)$ around $A$ such that $A′ \sim A$ for every $A′ \in B(A)$. [[Courtesy](https://economics.stackexchange.com/a/13134/33107)]  
> **username_0**: I am sorry although your answer was extremely helpful, I will have to accept the other answer as it clarifies a few more questions I had, including this.  
> **username_1**: @dictatemetokcus, no problem!  

---

