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).

2 Answers 2


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.

  • $\begingroup$ 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, 2018 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, 2018 at 3:12
  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 $x=y$.


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