<p>I use the usual notation $(\succeq,\succ,\sim)$ instead of $R$. I assume that $X_{\sim}$ in your question is the set of equivalence classes of $\sim$ on $X$.</p>

<p>Suppose that the cardinality of $X_{\sim}$ is equal to $n \geq 1$. There exist $x_1\cdots,x_n$ such that $x_1 \prec \cdots \prec x_n$. Define $v(x_1)=1,v(x_2)=2,\cdots,v (x_n)=n$. </p>

<p>Since $\sim$ has $n$ equivalence classes, for any $y \notin \{x_1,\cdots,x_n\}$ there exists a unique $x \in \{x_1,\cdots,x_n\}$ such that $y \sim x$. Define $v (y)=v(x)$.</p>

<p>The function $v$ defined on $X$ takes values in $\mathbb{N}$. I leave it to you to check that $v$ is a utility representation of $\succeq $.</p>

CC BY-SA 3.0
2017-03-30T06:56:04.583Z
2017-03-30T08:42:48.237Z
2017-03-30T08:42:48.237Z
Oliv
<p>The key is that $X_\sim$ is a finite set. That means that there exists an element $x_i \in X$ for which there is no element $x_j \in X$ where $x_j &lt; x_i$. Thus we can safely map $x_i$ (and all other elements that belong to the same equivalence class) to 0. We can then repeat the process with the remaining elements and map then next equivalence class to 1 and so on.</p>

<p>If $X_\sim$ had not been finite there would have been no guarantee of the existence of a smallest value in $X$.</p>

CC BY-SA 3.0
2017-03-30T08:46:16.940Z
$X$ is not assumed to be finite in OP's question. This reasoning only requires $X_{\sim}$ to be finite, which is possible even with an infinite $X$.
Oliv
CC BY-SA 3.0
2017-03-30T09:02:20.777Z
I am unfamiliar with the terminology. What is $X_\sim$? Is it $X$ but with equivalent (mutually uncomparable) elements collapsed into one?
Klas Lindbäck
CC BY-SA 3.0
2017-03-30T09:24:30.017Z
I think it is the set of equivalence classes of $\sim$, i.e. of subsets of $X$ that are ranked equally by $\sim$. For instance, if $X=\mathbb{R}$ and if $\succeq$ is defined by the utility function $v(x)=1$ if $x \geq 0$ and $v(x)=0$ otherwise, $X_{\sim}$ is simply composed of two subsets: $\mathbb{R}_+$ and its complement.
Oliv
CC BY-SA 3.0
2017-03-30T08:42:32.767Z
2017-03-30T10:06:51.517Z
2017-03-30T10:06:51.517Z
Klas Lindbäck
<p>How can we prove the following:  if a relation $R$ is a weak order on a set $X$ and $X_\sim$ is finite, then there exist a function $v:X\to \mathbb{N}$, which is a value representation of preference relation $R$.</p>

<p>Here I know the construction of the function $v$ when the range is the set of real numbers (as $X_\sim$ is countable). But,I am confused  when the range of $v$ is the set of natural numbers.</p>

<p>Could anyone please help me?</p>

CC BY-SA 3.0
2017-03-29T23:27:54.523Z
What $X_\sim$ represents ?
Alecos Papadopoulos
CC BY-SA 3.0
2017-03-29T23:49:06.607Z
Ok. And what is the difference between $X$ and  $X_\sim$?
Alecos Papadopoulos
CC BY-SA 3.0
2017-03-30T00:01:02.297Z
Are you sure that the elements of $X$ are each just one number? Maybe $X \subset \mathbb N^m$?
Alecos Papadopoulos
<microeconomics><preferences>
Value representation theorem for countable set
UserAb
# Value representation theorem for countable set

(Asked by: username_0 on 2017-03-29)

How can we prove the following: if a relation $R$ is a weak order on a set $X$ and $X\_\sim$ is finite, then there exist a function $v:X\to \mathbb{N}$, which is a value representation of preference relation $R$.

Here I know the construction of the function $v$ when the range is the set of real numbers (as $X\_\sim$ is countable). But,I am confused when the range of $v$ is the set of natural numbers.

Could anyone please help me?

**Tags:** microeconomics, preferences

> **Comments:**
> **username_1**: What $X_\sim$ represents ?  
> **username_1**: Ok. And what is the difference between $X$ and  $X_\sim$?  
> **username_1**: Are you sure that the elements of $X$ are each just one number? Maybe $X \subset \mathbb N^m$?  

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## 2 answers

### Answer by username_2 on 2017-03-30. Score: 1 | Accepted answer

I use the usual notation $(\succeq,\succ,\sim)$ instead of $R$. I assume that $X\_{\sim}$ in your question is the set of equivalence classes of $\sim$ on $X$.

Suppose that the cardinality of $X\_{\sim}$ is equal to $n \geq 1$. There exist $x\_1\cdots,x\_n$ such that $x\_1 \prec \cdots \prec x\_n$. Define $v(x\_1)=1,v(x\_2)=2,\cdots,v (x\_n)=n$.

Since $\sim$ has $n$ equivalence classes, for any $y \notin \{x\_1,\cdots,x\_n\}$ there exists a unique $x \in \{x\_1,\cdots,x\_n\}$ such that $y \sim x$. Define $v (y)=v(x)$.

The function $v$ defined on $X$ takes values in $\mathbb{N}$. I leave it to you to check that $v$ is a utility representation of $\succeq $.


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### Answer by username_3 on 2017-03-30. Score: 0

The key is that $X\_\sim$ is a finite set. That means that there exists an element $x\_i \in X$ for which there is no element $x\_j \in X$ where $x\_j < x\_i$. Thus we can safely map $x\_i$ (and all other elements that belong to the same equivalence class) to 0. We can then repeat the process with the remaining elements and map then next equivalence class to 1 and so on.

If $X\_\sim$ had not been finite there would have been no guarantee of the existence of a smallest value in $X$.

> **Comments:**
> **username_2**: $X$ is not assumed to be finite in OP's question. This reasoning only requires $X_{\sim}$ to be finite, which is possible even with an infinite $X$.  
> **username_3**: I am unfamiliar with the terminology. What is $X_\sim$? Is it $X$ but with equivalent (mutually uncomparable) elements collapsed into one?  
> **username_2**: I think it is the set of equivalence classes of $\sim$, i.e. of subsets of $X$ that are ranked equally by $\sim$. For instance, if $X=\mathbb{R}$ and if $\succeq$ is defined by the utility function $v(x)=1$ if $x \geq 0$ and $v(x)=0$ otherwise, $X_{\sim}$ is simply composed of two subsets: $\mathbb{R}_+$ and its complement.  

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