<p>On the one hand, every preference represented by a continuous utility function must be continuous. This follows from the fact that, for every $x \in X$, the "not-better-than" and "not-worse-than" sets can be written as
$$\left\{y\in X \middle| x \succsim y\right\} = u^{-1}\left(\left(-\infty,u(x)\right]\right),$$
and
$$\left\{y\in X \middle| y \succsim x\right\} = u^{-1}\left(\left[u(x),+\infty\right)\right),$$
where $u:X\to\mathbb{R}$ is any utility representation of $\succsim$. Then, if $u$ happens to be continuous, these sets have to be closed (as they are the inverse images of closed subsets of $\mathbb{R}$).</p>

<p>On the other hand, there are discontinuous preferences that admit a (discontinuous) utility representation. For an example, consider the (rational) preference on $[0,1]$ defined by $u(0)=0$, $u(x)=1$ for $x&gt;0$. This preference is not continuous, but it can obviously be represented (by $u$).</p>

CC BY-SA 4.0
2020-05-16T03:54:44.970Z
Hi! I'm trying to understand your answer. When you say "as they are the inverse images of closed subsets of $\mathbb{R}$". Who are these 'closed subsets'? $[u(x), +\infty)$?
Pedro Cunha
CC BY-SA 3.0
2018-02-25T13:18:43.023Z
2018-02-26T16:40:44.267Z
2018-02-26T16:40:44.267Z
john
<p>The easiest way to prove it is using the 'old' definition of continuity. 
$\succ$ is continuous iff whenever $x\succ y$, there exists neighborhoods of $x$ and $y$, $B_x, B_y$, such that all $z\in B(x)$ and $z'\in B(y)$, $z\succ z'$.</p>

<p>Suppose $x\succ y$. Because $u$ represents $\succ$, $u(x)&gt;u(y)$. Let $2\epsilon=u(x)-u(y)$. Because $u$ is continuous, there exists some $\delta&gt;0$ such that for all $z\in B_{\delta}(x)$, $u(z)&gt;u(x)-\epsilon$. Similarly, for all $z'\in B_{\delta}(y)$, $u(z')&gt;u(y)+\epsilon$. But then for all $z\in B_{\delta}(x)$ and $z'\in B_{\delta}(y)$, $z\succ z'$ as required.</p>

CC BY-SA 3.0
2017-09-11T07:54:57.763Z
2017-09-11T07:54:57.763Z
Fato
<p>I am trying to think of a preference relation that can be represented by a utility function but such that there does not exist a continuous utility function. </p>

<p>I know that you can represent continuous preferences with a discontinuous utility function, but I am not sure if the opposite is true. I am struggling to show that NO such continuous utility function exists.</p>

<p>My thoughts are that if you can define something with discontinuous preferences then maybe you can use this to imply that there do not exist any continuous utility functions. Note that Lexiographic preferences will not work because I am interested in a preference that can be represented by a utility function (albeit discontinuous).</p>

CC BY-SA 3.0
2017-09-11T03:25:09.737Z
A contrapositive may help: If preferences can be represented by a continuous utility function, does this imply that the preferences are continuous?
Ziwei Wang
<microeconomics><utility><preferences>
Do discontinuous preferences imply no continuous utility function?
user345
# Do discontinuous preferences imply no continuous utility function?

(Asked by: username_0 on 2017-09-10)

I am trying to think of a preference relation that can be represented by a utility function but such that there does not exist a continuous utility function.

I know that you can represent continuous preferences with a discontinuous utility function, but I am not sure if the opposite is true. I am struggling to show that NO such continuous utility function exists.

My thoughts are that if you can define something with discontinuous preferences then maybe you can use this to imply that there do not exist any continuous utility functions. Note that Lexiographic preferences will not work because I am interested in a preference that can be represented by a utility function (albeit discontinuous).

**Tags:** microeconomics, utility, preferences
> **Comments:**
> **username_1**: A contrapositive may help: If preferences can be represented by a continuous utility function, does this imply that the preferences are continuous?  

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## 2 answers

### Answer by username_4 on 2017-09-11. Score: 8 | Accepted answer

The easiest way to prove it is using the 'old' definition of continuity.
$\succ$ is continuous iff whenever $x\succ y$, there exists neighborhoods of $x$ and $y$, $B\_x, B\_y$, such that all $z\in B(x)$ and $z'\in B(y)$, $z\succ z'$.

Suppose $x\succ y$. Because $u$ represents $\succ$, $u(x)>u(y)$. Let $2\epsilon=u(x)-u(y)$. Because $u$ is continuous, there exists some $\delta>0$ such that for all $z\in B\_{\delta}(x)$, $u(z)>u(x)-\epsilon$. Similarly, for all $z'\in B\_{\delta}(y)$, $u(z')>u(y)+\epsilon$. But then for all $z\in B\_{\delta}(x)$ and $z'\in B\_{\delta}(y)$, $z\succ z'$ as required.


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### Answer by username_2 on 2018-02-25. Score: 4

On the one hand, every preference represented by a continuous utility function must be continuous. This follows from the fact that, for every $x \in X$, the "not-better-than" and "not-worse-than" sets can be written as
$$\left\{y\in X \middle| x \succsim y\right\} = u^{-1}\left(\left(-\infty,u(x)\right]\right),$$
and
$$\left\{y\in X \middle| y \succsim x\right\} = u^{-1}\left(\left[u(x),+\infty\right)\right),$$
where $u:X\to\mathbb{R}$ is any utility representation of $\succsim$. Then, if $u$ happens to be continuous, these sets have to be closed (as they are the inverse images of closed subsets of $\mathbb{R}$).

On the other hand, there are discontinuous preferences that admit a (discontinuous) utility representation. For an example, consider the (rational) preference on $[0,1]$ defined by $u(0)=0$, $u(x)=1$ for $x>0$. This preference is not continuous, but it can obviously be represented (by $u$).

> **Comments:**
> **username_3**: Hi! I'm trying to understand your answer. When you say "as they are the inverse images of closed subsets of $\mathbb{R}$". Who are these 'closed subsets'? $[u(x), +\infty)$?  

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