Consider the function $u : \mathbb{R}^2 \to \mathbb{R}$ that the following properties are satisfied:
Every level curve is a function from $\mathbb{R}$ to $\mathbb{R}$. (That is, a level curve $U(x,y)= c$ can be written as $y = f(x)$ for some $f : \mathbb{R} \to \mathbb{R}$.)
$\forall$ $p \in \mathbb{R}^2$, the sets $L(p)$ and $U(p)$ are closed. (Definition below.)
Definition: For $p \in \mathbb{R}^2$, define the lower contour of $p$ as $L(p) := \{(x,y) \in \mathbb{R}^2 : u(x,y) \leq u(p)\}$ and the upper contour of $p$ as $U(p) := \{(x,y) \in \mathbb{R}^2 : u(x,y) \geq u(p)\}$.
Question: Let $X = \mathbb{R}^2$. Given $u(\cdot, \cdot)$, can $u$ have a level curve $y = f(x)$ such that $f$ is discontinuous?
I posted this question originally on Economics SE that I could solve only partially. This is essentially the same question without the mention of preference (or the binary relation). Here's the original question:
Let $X = \mathbb{R}^2$. Suppose $\succeq$ denotes a continuous preference relation. If every indifference curve can be represented by functions from $\mathbb{R}$ to $\mathbb{R}$, will it mean the ICs will (necessarily) be continuous functions?
If you want to see my attempt (and/or post an answer), please click on this.
I think I have figured it out. Here's my new attempt :
Assume that there exists some $p = (x_0, y_0)$, $t = u(x_0, y_0)$, $f$ being the IC of $u = t$, such that $f$ is not continuous at $x_0$. You can find $(x_n)$ such that $x_n \to x_0$, and $f(x_n)$ does not converge to $y_0$.
Up to extracting a subsequence, you can assume that $f(x_n)$ converges to some element in $\mathbb{R} \cup \{ +\infty, -\infty \}$. Denote this limit by $y'$.
If $y'$ is finite, since $(x_n, f(x_n))$ are in both $L(p)$ and $U(p)$ and these sets are closed, $(x_0, y')$ as well, and therefore $u(x_0, y') = u(x_0, y_0)$. But $y_0 \neq y'$ and the level set is represented by $f$, this is a contradiction.
Now change your point of view and consider, at fixed $x$, the values taken by $u(x, y)$ when $y$ varies, $x$ is fixed. $y \mapsto u(x, y)$ is injective by the fact that each level set can be represented by a function. It is even bijective since, for each $c$, there exists $f_c$ and $y = f_c(x)$ such that $u(x, y) = c$. Furthermore, $\{ y, u(x,y) \geq u(x, a) \}$ is closed for each fixed $a$, $x$, so the sets $\{ y, u(x, y) \geq c \}$ are closed for every fixed $c$, $x$. The same holds if you reverse the inequality, so $y \mapsto u(x, y)$ is continuous for each $x$. Being bijective, it is a monotone map.
Let's return to $y'$ when it is infinite and assume, for instance, $y' = +\infty$ (up to a symmetry on the whole plane $\mathbb{R}^2$, you can always assume $y' = +\infty$). For each $n$, $y \mapsto u(x_n, y)$ is either increasing or decreasing : but, since there are either infinitely many $n$ such that it is increasing or infinitely many such that it is decreasing, you may assume up to extracting a subsequence that they are all increasing or all decreasing. Likewise, $y \mapsto u(x_0, y)$ is also either increasing or decreasing.
For $n$ large enough, $f(x_n) > y_0+1$ and thus $(x_n, y_0-1)$ and $(x_n, y_0+1)$ are both elements of $U(p)$ or $L(p)$ (depending on whether $y \mapsto u(x_n, y)$ are increasing or decreasing). Both sets are closed, so $(x_0, y_0+1)$ and $(x_0, y_0-1)$ are both elements of the corresponding set. But this contradicts the (strict) monotonicity of $y \mapsto u(x_0, y)$, since we know the value $u(x_0, y_0)$.