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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.

Rainroad
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  • Related: https://economics.stackexchange.com/a/12411/42 – Herr K. Sep 27 '19 at 04:55
  • 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 05:27
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    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 01:02
  • 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 01:28
  • Yeah that's true, but then in the exercises sometimes it is specified clearly that u(x) is continuous and sometimes it is not, so I get kinda confused on that. – Rainroad Sep 28 '19 at 22:35
  • And also in the solutions, the exact phrasing was "X is closed because of $u(x) \geq \bar{u}$ and $x \in R^L_+$" which doesn't seem to necessitate the continuity condition to me. – Rainroad Sep 28 '19 at 22:36

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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.

Giskard
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