<p>The statement does not seem to be true.</p>

<p>Define <span class="math-container">$u$</span> as <span class="math-container">$u(x) = -1$</span> if <span class="math-container">$x \leq 0$</span>, <span class="math-container">$u(x) = 0$</span> if <span class="math-container">$x \in (0,1)$</span> and <span class="math-container">$u(x) = 1$</span> if <span class="math-container">$x \geq 1$</span>. The set of points <span class="math-container">$x$</span> for which <span class="math-container">$u(x) \geq 0$</span> is <span class="math-container">$(0,\infty)$</span>, which is not closed.</p>

CC BY-SA 4.0
2019-10-31T19:24:45.543Z
@domotorp If $u$ is a utility function then universal domain and monotonicity are usual assumptions.
Giskard
CC BY-SA 4.0
2019-11-01T07:45:35.240Z
@domotorp It could be.
Giskard
CC BY-SA 4.0
2019-09-27T05:50:15.797Z
2019-09-27T05:50:15.797Z
Giskard
<p>Prove that the set <span class="math-container">$X = \{x \in R^L_+| u(x) \geq \bar u\}$</span> is closed. </p>

<p>Saw this statement in the textbook but I'm not sure how this is the case when we don't have any restrictions on <span class="math-container">$u(x)$</span> such as continuity. I can prove this if it is continuous, but I'm not sure how to do it if  isn't. </p>

CC BY-SA 4.0
2019-09-27T04:55:57.927Z
Related: https://economics.stackexchange.com/a/12411/42
Herr K.
CC BY-SA 4.0
2019-09-27T05:27:09.073Z
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
CC BY-SA 4.0
2019-09-27T19:26:20.710Z
Are you sure you're not leaving out any context? Which textbook are you using?
Herr K.
CC BY-SA 4.0
2019-09-28T01:02:10.083Z
MGW but the solutions manual, which is not written by the authors of the book.
Rainroad
CC BY-SA 4.0
2019-09-28T01:28:49.540Z
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.
CC BY-SA 4.0
2019-09-28T22:35:10.650Z
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
CC BY-SA 4.0
2019-09-28T22:36:40.370Z
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
<microeconomics><mathematical-economics>
Prove that the set $X = \{x \in R^L_+| u(x) \geq \bar u\}$ is closed
Rainroad
# Prove that the set $X = \{x \in R^L_+| u(x) \geq \bar u\}$ is closed

(Asked by: username_0 on 2019-09-27)

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.

**Tags:** microeconomics, mathematical-economics
> **Comments:**
> **username_1**: Related: https://economics.stackexchange.com/a/12411/42  
> **username_0**: 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.  
> **username_1**: Are you sure you're not leaving out any context? Which textbook are you using?  
> **username_0**: MGW but the solutions manual, which is not written by the authors of the book.  
> **username_1**: 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.  
> **username_0**: 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.  
> **username_0**: 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.  

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## 1 answers

### Answer by username_2 on 2019-09-27. Score: 1

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.

> **Comments:**
> **username_2**: @domotorp If $u$ is a utility function then universal domain and monotonicity are usual assumptions.  
> **username_2**: @domotorp It could be.  

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