<p>For a set to be continuous, it's contour sets must be closed. Since we can define <span class="math-container">$$ \succsim x^0 = \{x, x^0 \in X: x \succsim x^0 \} $$</span> and <span class="math-container">$$ \precsim x^0 = \{x, x^0 \in X: x \precsim x^0\}$$</span></p>

<p>it can be seen that <span class="math-container">$\succsim x^0$</span> is the upper contour set of <span class="math-container">$X$</span> and <span class="math-container">$\precsim x^0$</span> is the lower contour set of <span class="math-container">$X$</span>. Therefore, if <span class="math-container">$\succsim$</span> is continuous on <span class="math-container">$X$</span>, it must the case that both <span class="math-container">$\succsim x^0$</span> and <span class="math-container">$\precsim x^0$</span> are closed sets.</p>

<p>Is my answer correct?</p>

<p>Edit: definition of continuous preference I'm using:</p>

<p><a href="https://i.sstatic.net/MFtcs.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MFtcs.png" alt="Continuous preference - MGW"></a></p>

CC BY-SA 4.0
2020-05-13T03:31:47.873Z
The argument looks circular to me. What is your definition of continuous preferences?
Regio
CC BY-SA 4.0
2020-05-13T03:47:17.000Z
I'll add it to the question, @Regio
Pedro Cunha
CC BY-SA 4.0
2020-05-13T04:02:04.670Z
Then your answer is not correct. You claim "For a set to be continuous, it's contour sets must be closed", but provide no proof of such a statement. Note that being able to write the definition of the upper contour sets and lower contour sets of $x_0$ is not a sufficient argument to show they are closed.
Regio
CC BY-SA 4.0
2020-05-13T04:03:30.600Z
I see. I thought I could simply use the definition provided by MGW and start from there. If it's not too much trouble, could you point me in the right direction of the proof?
Pedro Cunha
CC BY-SA 4.0
2020-05-13T04:08:55.123Z
MGW is almost spelling out the complete proof,  you want to take a Cauchy sequence inside the set $\succsim x^0$ and show that its limit point is also in the set. Remember that a set is closed if it contains all its limit points.
Regio
CC BY-SA 4.0
2020-05-13T11:38:27.550Z
I guess it does, @KennyLJ. I'm sorry for asking something rather obvious, it's just that it's my first time dealing with MGW and it's been somewhat overwhelming. I appreciate the tips and candor. I'll read it until I can understand the proof. Thanks!
Pedro Cunha
<preferences>
Show that if $\succsim$ is continous on $X$, then the sets $ \precsim (x^0)$ and $\succsim (x^0)$ are closed
Pedro Cunha
# Show that if $\succsim$ is continous on $X$, then the sets $ \precsim (x^0)$ and $\succsim (x^0)$ are closed

(Asked by: username_0 on 2020-05-13)

For a set to be continuous, it's contour sets must be closed. Since we can define $$ \succsim x^0 = \{x, x^0 \in X: x \succsim x^0 \} $$ and $$ \precsim x^0 = \{x, x^0 \in X: x \precsim x^0\}$$

it can be seen that $\succsim x^0$ is the upper contour set of $X$ and $\precsim x^0$ is the lower contour set of $X$. Therefore, if $\succsim$ is continuous on $X$, it must the case that both $\succsim x^0$ and $\precsim x^0$ are closed sets.

Is my answer correct?

Edit: definition of continuous preference I'm using:

[![Continuous preference - MGW](https://i.sstatic.net/MFtcs.png)](https://i.sstatic.net/MFtcs.png)

**Tags:** preferences
> **Comments:**
> **username_1**: The argument looks circular to me. What is your definition of continuous preferences?  
> **username_0**: I'll add it to the question, @Regio  
> **username_1**: Then your answer is not correct. You claim "For a set to be continuous, it's contour sets must be closed", but provide no proof of such a statement. Note that being able to write the definition of the upper contour sets and lower contour sets of $x_0$ is not a sufficient argument to show they are closed.  
> **username_0**: I see. I thought I could simply use the definition provided by MGW and start from there. If it's not too much trouble, could you point me in the right direction of the proof?  
> **username_1**: MGW is almost spelling out the complete proof,  you want to take a Cauchy sequence inside the set $\succsim x^0$ and show that its limit point is also in the set. Remember that a set is closed if it contains all its limit points.  
> **username_0**: I guess it does, @KennyLJ. I'm sorry for asking something rather obvious, it's just that it's my first time dealing with MGW and it's been somewhat overwhelming. I appreciate the tips and candor. I'll read it until I can understand the proof. Thanks!  

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