3
$\begingroup$

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. $\blacksquare$.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – VARulle
    Jul 8 at 10:01
  • $\begingroup$ @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? $\endgroup$ Jul 8 at 10:38
  • 3
    $\begingroup$ @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? $\endgroup$
    – tdm
    Jul 8 at 11:15
  • $\begingroup$ @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. $\endgroup$ Jul 8 at 11:55
7
$\begingroup$

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 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 $\square$.

$\endgroup$
2
  • $\begingroup$ 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 $I$. $\endgroup$ Jul 9 at 1:58
  • $\begingroup$ 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. $\endgroup$ Jul 9 at 2:00
7
$\begingroup$

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. 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.

$\endgroup$
4
  • $\begingroup$ 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.) $\endgroup$ Jul 9 at 2:03
  • $\begingroup$ 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] $\endgroup$ Jul 9 at 2:05
  • $\begingroup$ 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. $\endgroup$ Jul 9 at 2:07
  • 1
    $\begingroup$ @dictatemetokcus, no problem! $\endgroup$
    – VARulle
    Jul 9 at 7:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.