0
$\begingroup$

Can anybody give an intuitively explanation for the following problem?

Let $\succeq$ be a preference relation on a set X. Define I(x) to be the set of all y ∈ X for which y ∼ x. Show that the set (of sets!) {I(x)|x ∈ X} is a partition of X, i.e.,

  • For all x and y, either I(x) = I(y) or I(x) ∩ I(y) = ∅.
  • For every x ∈ X, there is y ∈ X such that x ∈ I(y).
$\endgroup$
0
$\begingroup$

Let $X$ be the set over which a preference $\succsim$ is defined. The question asks you to prove that $X$ can be divided up into numerous subsets defined by the indifference relation, i.e. all elements within an indifference set, denoted $I(x)$, have the same desirability according to the preference $\succsim$. The collection of these indifference sets is denoted $\{I(x)\vert x\in X\}$. The two bullet points tell you how to proceed with the proof.

First, you need to argue that the indifference sets defined by any two elements $x$ and $y$ are either identical or distinct.

Second, you need to argue that every element belongs to one of the indifference sets that divide up $X$.


Here is a simplified example to give you some intuition.

Suppose the set $X$ contains the following items: an apple ($a$), a banana ($b$), a cherry ($c$), a dragon fruit ($d$); that is, $X=\{a,b,c,d\}$. I like cherries best, bananas and apples the second (I'm also indifferent between the two), and dragon fruits the least. This is my preference over $X$. Define the indifference sets as follows:

  • $I(a)=\{a,b\}$ (i.e. the set that contains items indifferent to apple includes both apple and banana);
  • $I(b)=\{a,b\}$;
  • $I(c)=\{c\}$; and
  • $I(d)=\{d\}$.

As we can see, for any $x,y\in\{a,b,c,d\}$, $I(x)$ and $I(y)$ are either identical (as when $x=a$ and $y=b$, or vice versa) or distinct (as when at least one of $x$ and $y$ is $c$ or $d$). This satisfies the first requirement.

For the second requirement, it is clear that the four items ($a,b,c$ and $d$) belongs to one of the three indifference sets ($\{a,b\},\{c\},\{d\}$).

To properly prove the claim, you need to generalize the example above to cases where the set $X$ contains an arbitrary number of elements. Strictly speaking, you'd also need to consider arbitrary preferences. But since the preference relation $\succsim$ is complete and transitive, such consideration is less crucial.

$\endgroup$
2
  • $\begingroup$ Could you also give me a hint how to do the proof? Or recommend literature? I still have problems in the topic of preference relations... $\endgroup$ Jul 3 '18 at 6:42
  • $\begingroup$ @hermanzegerman: I added an example. Hopefully it'll give you some intuition on how to construct a proof. $\endgroup$
    – Herr K.
    Jul 5 '18 at 3:12
0
$\begingroup$
  1. Either $I(x)=I(y)$ or $I(x) \cap I(y)=\phi $ $\forall$ $x,y\in X$

Proof by contradiction

Let there be $x,y \in X$ such that $x\neq y$ and $I(x)\neq I(y)$. Let there be an element $a \in I(x)\cap I(y)$

Hence by definition of $I(.)$, $x\sim a$ and $y\sim a$ which implies that $x\sim y$ and this implies that $I(x)=I(y)$ which is a contradiction.

  1. $\forall x\in X, \exists y\in X, such\ that\ x\in I(y)$

Every element $x\in X$ has the property that $x\sim x$, hence there is at least one element in $X$, $\forall y\in X$ such that $x=I(y)$, which is the case when $x=y$.

$\endgroup$

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.