<p>Giskard is right lexicographic preferences would make the job being dicsontinuous, but the problem is to find relations with closed contour sets. Maybe an idea would be to start from constructing a preference relation  for which the only possible upper and lower contour sets are <span class="math-container">$X$</span> and <span class="math-container">$\emptyset$</span>. In that case, you take the indiscrete topology, and all nets (or generalized sequences) converge to all the points in <span class="math-container">$X$</span>. Anyway, I have not been able to construct such a relation yet.</p>

CC BY-SA 4.0
2020-03-24T12:08:30.277Z
2020-03-24T12:08:30.277Z
giusky
<p>Let <span class="math-container">$\succsim$</span> be a transitive and reflexive relation on a metric space <span class="math-container">$X$</span> with closed upper and lower contour sets. If <span class="math-container">$\succsim$</span> is not complete, does it hold that: for all converging sequences with <span class="math-container">$x_n\succsim y_n$</span> for each <span class="math-container">$n\geq 1$</span> and <span class="math-container">$x_n\to x$</span>, <span class="math-container">$y_n\to y$</span>, we have <span class="math-container">$x\succsim y$</span>? I think that without the completeness of <span class="math-container">$\succsim$</span> this cannot hold, but I failed to provide an example.  </p>

CC BY-SA 4.0
2020-03-24T11:56:19.483Z
If $X$ is not complete that means there are $x$ and $y$ such that neither $x \succeq y$ nor $x \preceq y$ hold. You can easily fabricate an example where $x_n \to x$ and $y_n \to y$ while for all $n$ you have $x_n \preceq y_n$. Then all you have to show is that the contour sets are indeed closed, right?
Giskard
CC BY-SA 4.0
2020-03-24T12:42:35.030Z
Yes, that's what I am struggling to show. I can't find a relation with closed contour sets, which is not "continuous" in terms of sequences.
grintaaaaaaa
CC BY-SA 4.0
2020-03-24T23:40:38.807Z
A result by Schmeidler (1971) tells us if $X$ is a connected topological space with a transitive, continuous (i.e. upper contour sets are closed and strict upper contour sets are open) and non-trivial (i.e. not every element is indifferent to one another), it is a complete preference. You must break one of these requirements for a counterexample. (e.g. using the discrete topology as per the answer).
Walrasian Auctioneer
<microeconomics><mathematical-economics><preferences><decision-theory><theory>
Continuity of preferences
grintaaaaaaa
# Continuity of preferences

(Asked by: username_0 on 2020-03-24)

Let $\succsim$ be a transitive and reflexive relation on a metric space $X$ with closed upper and lower contour sets. If $\succsim$ is not complete, does it hold that: for all converging sequences with $x\_n\succsim y\_n$ for each $n\geq 1$ and $x\_n\to x$, $y\_n\to y$, we have $x\succsim y$? I think that without the completeness of $\succsim$ this cannot hold, but I failed to provide an example.

**Tags:** microeconomics, mathematical-economics, preferences, decision-theory, theory
> **Comments:**
> **username_1**: If $X$ is not complete that means there are $x$ and $y$ such that neither $x \succeq y$ nor $x \preceq y$ hold. You can easily fabricate an example where $x_n \to x$ and $y_n \to y$ while for all $n$ you have $x_n \preceq y_n$. Then all you have to show is that the contour sets are indeed closed, right?  
> **username_0**: Yes, that's what I am struggling to show. I can't find a relation with closed contour sets, which is not "continuous" in terms of sequences.  
> **username_2**: A result by Schmeidler (1971) tells us if $X$ is a connected topological space with a transitive, continuous (i.e. upper contour sets are closed and strict upper contour sets are open) and non-trivial (i.e. not every element is indifferent to one another), it is a complete preference. You must break one of these requirements for a counterexample. (e.g. using the discrete topology as per the answer).  

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## 1 answers

### Answer by username_3 on 2020-03-24. Score: 0 | Accepted answer

Giskard is right lexicographic preferences would make the job being dicsontinuous, but the problem is to find relations with closed contour sets. Maybe an idea would be to start from constructing a preference relation for which the only possible upper and lower contour sets are $X$ and $\emptyset$. In that case, you take the indiscrete topology, and all nets (or generalized sequences) converge to all the points in $X$. Anyway, I have not been able to construct such a relation yet.


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