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).

  • 2
    $\begingroup$ A contrapositive may help: If preferences can be represented by a continuous utility function, does this imply that the preferences are continuous? $\endgroup$
    – Ziwei Wang
    Commented Sep 11, 2017 at 3:25

2 Answers 2


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.


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 $u$).

  • $\begingroup$ 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'? $[u(x), +\infty)$? $\endgroup$ Commented May 16, 2020 at 3:54

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