# Prove that the set $X = \{x \in R^L_+| u(x) \geq \bar u\}$ is closed

Prove that the set $$X = \{x \in R^L_+| u(x) \geq \bar u\}$$ is closed.

Saw this statement in the textbook but I'm not sure how this is the case when we don't have any restrictions on $$u(x)$$ such as continuity. I can prove this if it is continuous, but I'm not sure how to do it if isn't.

• Sep 27 '19 at 4:55
• Yeah but it doesn't quite answer my question. I don't think this is true unless u () is continuous. The book says that this is closed due to $u(x) \geq \bar{u}$ and $x \in R^L_+$ but that doesn't seem quite true to me. This statement implies that the upper contour set is always closed no matter what the preference is, but this can't be true. Sep 27 '19 at 5:27
• Are you sure you're not leaving out any context? Which textbook are you using? Sep 27 '19 at 19:26
• MGW but the solutions manual, which is not written by the authors of the book. Sep 28 '19 at 1:02
• At the beginning of Section 3.D, MWG do make a few assumptions that affect the rest of the chapter, and $u(x)$ being continuous is one of them. Sep 28 '19 at 1:28

Define $$u$$ as $$u(x) = -1$$ if $$x \leq 0$$, $$u(x) = 0$$ if $$x \in (0,1)$$ and $$u(x) = 1$$ if $$x \geq 1$$. The set of points $$x$$ for which $$u(x) \geq 0$$ is $$(0,\infty)$$, which is not closed.
• Why did you need the $u(x)=1$ part? Oct 31 '19 at 19:17
• @domotorp If $u$ is a utility function then universal domain and monotonicity are usual assumptions. Oct 31 '19 at 19:24