<p>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. </p>

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

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

<hr>

<p>Here is a simplified example to give you some intuition. </p>

<p>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: </p>

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

<p>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. </p>

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

<p>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. </p>

CC BY-SA 4.0
2018-07-03T06:42:52.297Z
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...
hermanzegerman
CC BY-SA 4.0
2018-07-05T03:12:45.623Z
@hermanzegerman: I added an example. Hopefully it'll give you some intuition on how to construct a proof.
Herr K.
CC BY-SA 4.0
2018-07-02T16:54:31.370Z
2018-07-05T03:11:48.373Z
2018-07-05T03:11:48.373Z
Herr K.
<ol>
<li>Either $I(x)=I(y)$ or $I(x) \cap I(y)=\phi $ $\forall$ $x,y\in X$</li>
</ol>

<p><strong>Proof by contradiction</strong></p>

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

<p>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.</p>

<ol start="2">
<li>$\forall x\in X, \exists y\in X, such\  that\  x\in I(y)$</li>
</ol>

<p>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 $x=y$.</p>

CC BY-SA 4.0
2018-07-05T04:06:01.903Z
2018-07-05T04:06:01.903Z
Saurav Goyal
<p>Can anybody give an intuitively explanation for the following problem?</p>

<blockquote>
  <p>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.,</p>
  
  <ul>
  <li>For all x and y, either I(x) = I(y) or I(x) ∩ I(y) = ∅. </li>
  <li>For every x ∈ X, there is y ∈ X such that x ∈ I(y).</li>
  </ul>
</blockquote>

<microeconomics><preferences>
preference relations
hermanzegerman
# preference relations

(Asked by: username_0 on 2018-07-02)

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).

**Tags:** microeconomics, preferences


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## 2 answers

### Answer by username_1 on 2018-07-02. Score: 0 | Accepted answer

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.

> **Comments:**
> **username_0**: 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...  
> **username_1**: @username_0: I added an example. Hopefully it'll give you some intuition on how to construct a proof.  

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### Answer by username_2 on 2018-07-05. Score: 0

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.

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

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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