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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.

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  • $\begingroup$ Related: economics.stackexchange.com/a/12411/42 $\endgroup$ – Herr K. Sep 27 at 4:55
  • $\begingroup$ 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. $\endgroup$ – Rainroad Sep 27 at 5:27
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    $\begingroup$ Are you sure you're not leaving out any context? Which textbook are you using? $\endgroup$ – Herr K. Sep 27 at 19:26
  • $\begingroup$ MGW but the solutions manual, which is not written by the authors of the book. $\endgroup$ – Rainroad Sep 28 at 1:02
  • $\begingroup$ 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. $\endgroup$ – Herr K. Sep 28 at 1:28
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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.

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  • $\begingroup$ Why did you need the $u(x)=1$ part? $\endgroup$ – domotorp Oct 31 at 19:17
  • $\begingroup$ @domotorp If $u$ is a utility function then universal domain and monotonicity are usual assumptions. $\endgroup$ – Giskard Oct 31 at 19:24
  • $\begingroup$ I mean why can't that be also 0? $\endgroup$ – domotorp Nov 1 at 5:38
  • $\begingroup$ @domotorp It could be. $\endgroup$ – Giskard Nov 1 at 7:45

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