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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 be continuous functions?

(In other words, will the level curves, if representable by functions from $\mathbb{R}$ to $\mathbb{R}$, of the utility function of $\succeq$, be continuous functions?)

Here are two proofs that work for jump and removable discontinuities:

Proof 1: Assume on the contrary. Consider any indifference curve $U(x,y) = c$ with a point of discontinuity at $x = t$. Denote the curve by $y = f(x)$. Let $x_n = t - \frac{\epsilon}{n}$ and $y_n = f(t - \frac{\epsilon}{n})$ such that $(x_n, y_n)$ converges to $L$. (WLOG, this assumes $\displaystyle\lim_{x \to t^{-}} f(x) \neq f(t)$.) Note that $L$ belongs to exactly one of the lower and the upper contours of a point $p$ such that $U(p) = c$ although it is a limit point of both the contours. Thus, both the upper/lower contours of $p$ can not be closed at the same time which contradicts the continuity of $\succeq$.

Proof 2: Let $x$ be the $L$ we constructed earlier, and $y$ be a point on the IC. Then (WLOG) $x \succ y$ but any open ball $B_x$ we construct will contain an element that has the same utility as of $y$ (as the ball will intersect with the IC).

There are several other types of essential discontinuities which do not get covered by the proofs. In particular, I can't show if $\succeq_{\text{cont.}}$ can have a discontinuous IC given the ICs are representable by functions from $\mathbb{R}$ to $\mathbb{R}$.

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  • $\begingroup$ What is your question? $\endgroup$
    – Giskard
    Apr 14, 2022 at 10:59
  • $\begingroup$ @Giskard To prove the statement in quotes. My idea works for jump and removable discontinuities (of the IC), but not for all types of essential discontinuities. $\endgroup$ Apr 14, 2022 at 12:53
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    $\begingroup$ I don't know the answer, but if a counterexample exists, it must involve functions that are not locally bounded. There must be a point such that the function is not bounded on any neighborhood. Otherwise, since indifference curves for continuous preferences are closed, the graph of these functions must be closed. If they are locally bounded, they locally take values in a compact space. The closed graph theorem guarantees then that they are continuous. $\endgroup$ Apr 17, 2022 at 16:30
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    $\begingroup$ It is also worth noting that if a proof exists, it must use the fact that all indifference curves are the graphs of functions, not just a specific one. An indifference curve for some continuous preference relation might be the graph of a discontinuous function: Let $C=\{(x,y)\mid x\neq y, y=1/x\}\cup\{(0,0)\}$ and define the utility function $v$ by letting $v(x,y)$ be the Euclidean distance between $(x,y)$ and $C$. The resulting utility function is continuous, but the indifference curve corresponding to a utility level of $0$ is $C$, the graph of a discontinuous function. $\endgroup$ Apr 22, 2022 at 13:08
  • $\begingroup$ @MichaelGreinecker That's a nice point. For those who come after, note that $v$ does not have the property that all indifference curves are graphs of functions. To see this note that $(0,1)$ is equidistant $(1,1)$ on the graph of $1/x$ and $(0,0)$. As you proceed from $(0,1)$ down the $y$-axis you take on every utility level in $[0,1]$ and as you proceed up from $(0,1)$ $v$ ranges over $(0,1]$. Hence you have many indifference curves failing to pass the vertical line test at $x=0$. $\endgroup$ Aug 17, 2022 at 21:09

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