<ol>
<li>If we know where <span class="math-container">$x_n$</span> stands with respect to the previous points, we need to give <span class="math-container">$U(x_n)$</span> a value that is higher than the corresponding values to its less preferred points until now, and a lower value than the corresponding values to its more preferred points until now. It has to be this way because utility functions are increasing with respect to preferences (monotone).</li>
<li><span class="math-container">$U(x_1)$</span> doesn’t need to be <span class="math-container">$\frac{1}{2}$</span>. We just need to set it somewhere in the interval as the base case of the inductive process.</li>
</ol>

CC BY-SA 4.0
2022-11-15T19:38:36.757Z
Everything is clear to me now. Thanks.
homo-economitux
CC BY-SA 4.0
2022-11-15T17:46:51.317Z
2022-11-15T20:57:36.377Z
2022-11-15T20:57:36.377Z
Nicolas Torres
<p><strong>Theorem:</strong> If <span class="math-container">$X$</span> is countable, then any preference on <span class="math-container">$X$</span> has a utility representation with a range <span class="math-container">$(0,1)$</span>.</p>
<p>The stated proof in Rubinstein's lecture notes:</p>
<p><strong>Proof:</strong> Let <span class="math-container">$\{x_n\}$</span> be an enumeration of all elements in <span class="math-container">$X$</span>. We will construct the utility function inductively. Set <span class="math-container">$U(x_1) =1/2$</span>. Assume that you have completed the definition of the values <span class="math-container">$U(x_1), . . ., U(x_{n-1})$</span> so that <span class="math-container">$x_k \succeq x_l$</span> iff <span class="math-container">$U(x_k) \geq U(x_l)$</span>. If <span class="math-container">$x_n$</span> is indifferent to <span class="math-container">$x_k$</span> for some <span class="math-container">$k&lt;n$</span>, then assign <span class="math-container">$U(x_n)=U(x_k)$</span>. If not, choose <span class="math-container">$U(x_n)$</span> to be between the two nonempty sets <span class="math-container">$\{U(x_k) | x_k \prec x_n \} \cup \{0\}$</span> and <span class="math-container">$\{U(x_k) | x_n \prec x_k \} \cup \{1\}$</span>. This is possible since by transitivity all numbers in the first set are below all numbers in the second set. Thus, for any <span class="math-container">$k&lt;n$</span> we have <span class="math-container">$x_n \succeq x_k$</span> iff <span class="math-container">$U(x_n) \geq U(x_k)$</span> and the function <span class="math-container">$U$</span> extended to <span class="math-container">$\{x_1, ..., x_n\}$</span> represents the preferences on those elements.
To complete the proof that <span class="math-container">$U$</span> represents <span class="math-container">$\succeq$</span>, take any two elements, <span class="math-container">$x$</span> and <span class="math-container">$y\in X$</span>. For some <span class="math-container">$k$</span> and <span class="math-container">$l$</span> we have <span class="math-container">$x=x_k$</span> and <span class="math-container">$y=y_l$</span>. The above applied to <span class="math-container">$n=\max\{k,l\}$</span> yields <span class="math-container">$x_k \succeq x_l$</span> iff <span class="math-container">$U(x_k) \geq U(x_l)$</span>.</p>
<p>I highlighted below the step in the proof I couldn't figure out. I tried to draw a number line or break the parts of the statement, yet couldn't understand.</p>
<blockquote>
<p>If not, choose <span class="math-container">$U(x_n)$</span> to be between the two nonempty sets <span class="math-container">$\{U(x_k)
&gt; | x_k \prec x_n \} \cup \{0\}$</span> and <span class="math-container">$\{U(x_k) | x_n \prec x_k \} \cup
&gt; \{1\}$</span>. This is possible since by transitivity all numbers in the
first set are below all numbers in the second set.</p>
</blockquote>
<p>Furthermore, why have we done this step? Is it necessary to complete the proof?</p>
<blockquote>
<p>Set <span class="math-container">$U(x_1) =1/2$</span>.</p>
</blockquote>
<p>Thanks for your help.</p>

CC BY-SA 4.0
2022-11-15T17:52:11.277Z
Are you familiar with [mathematical induction](https://en.wikipedia.org/wiki/Mathematical_induction#Description)? Setting $U(x_1)=1/2$ is the base case, and assigning value to $U(x_n)$, as described in your first quote, is the inductive step.
Herr K.
CC BY-SA 4.0
2022-11-15T19:37:34.257Z
I have some familiarity but I believe I need to learn more. I'll check the link you attached and also do some research. If you have any suggestions to study, I'll be more than glad. Thanks.
homo-economitux
<microeconomics><utility><preferences>
How to prove that any preference relation on (countable) X has a utility representation with a range (0,1)?
homo-economitux
# How to prove that any preference relation on (countable) X has a utility representation with a range (0,1)?

(Asked by: username_0 on 2022-11-15)

**Theorem:** If $X$ is countable, then any preference on $X$ has a utility representation with a range $(0,1)$.

The stated proof in Rubinstein's lecture notes:

**Proof:** Let $\{x\_n\}$ be an enumeration of all elements in $X$. We will construct the utility function inductively. Set $U(x\_1) =1/2$. Assume that you have completed the definition of the values $U(x\_1), . . ., U(x\_{n-1})$ so that $x\_k \succeq x\_l$ iff $U(x\_k) \geq U(x\_l)$. If $x\_n$ is indifferent to $x\_k$ for some $k, then assign $U(x\_n)=U(x\_k)$. If not, choose $U(x\_n)$ to be between the two nonempty sets $\{U(x\_k) | x\_k \prec x\_n \} \cup \{0\}$ and $\{U(x\_k) | x\_n \prec x\_k \} \cup \{1\}$. This is possible since by transitivity all numbers in the first set are below all numbers in the second set. Thus, for any $k we have $x\_n \succeq x\_k$ iff $U(x\_n) \geq U(x\_k)$ and the function $U$ extended to $\{x\_1, ..., x\_n\}$ represents the preferences on those elements.
To complete the proof that $U$ represents $\succeq$, take any two elements, $x$ and $y\in X$. For some $k$ and $l$ we have $x=x\_k$ and $y=y\_l$. The above applied to $n=\max\{k,l\}$ yields $x\_k \succeq x\_l$ iff $U(x\_k) \geq U(x\_l)$.

I highlighted below the step in the proof I couldn't figure out. I tried to draw a number line or break the parts of the statement, yet couldn't understand.

> If not, choose $U(x\_n)$ to be between the two nonempty sets $\{U(x\_k)
> > | x\_k \prec x\_n \} \cup \{0\}$ and $\{U(x\_k) | x\_n \prec x\_k \} \cup
> > \{1\}$. This is possible since by transitivity all numbers in the
> first set are below all numbers in the second set.

Furthermore, why have we done this step? Is it necessary to complete the proof?

> Set $U(x\_1) =1/2$.

Thanks for your help.

**Tags:** microeconomics, utility, preferences
> **Comments:**
> **username_1**: Are you familiar with [mathematical induction](https://en.wikipedia.org/wiki/Mathematical_induction#Description)? Setting $U(x_1)=1/2$ is the base case, and assigning value to $U(x_n)$, as described in your first quote, is the inductive step.  
> **username_0**: I have some familiarity but I believe I need to learn more. I'll check the link you attached and also do some research. If you have any suggestions to study, I'll be more than glad. Thanks.  

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## 1 answers

### Answer by username_2 on 2022-11-15. Score: 1 | Accepted answer

1. If we know where $x\_n$ stands with respect to the previous points, we need to give $U(x\_n)$ a value that is higher than the corresponding values to its less preferred points until now, and a lower value than the corresponding values to its more preferred points until now. It has to be this way because utility functions are increasing with respect to preferences (monotone).
2. $U(x\_1)$ doesn’t need to be $\frac{1}{2}$. We just need to set it somewhere in the interval as the base case of the inductive process.

> **Comments:**
> **username_0**: Everything is clear to me now. Thanks.  

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