# 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.

• – Herr K. 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. – Rainroad Sep 27 '19 at 5:27
• Are you sure you're not leaving out any context? Which textbook are you using? – Herr K. Sep 27 '19 at 19:26
• MGW but the solutions manual, which is not written by the authors of the book. – Rainroad 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. – Herr K. Sep 28 '19 at 1:28

The statement does not seem to be true.

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? – domotorp Oct 31 '19 at 19:17
• @domotorp If $u$ is a utility function then universal domain and monotonicity are usual assumptions. – Giskard Oct 31 '19 at 19:24
• I mean why can't that be also 0? – domotorp Nov 1 '19 at 5:38
• @domotorp It could be. – Giskard Nov 1 '19 at 7:45