New answers tagged preferences
0
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 ...
2
You cannot prove this. It is wrong.
Define $u(x)=\min{\{x,0\}}$. Let $\succsim$ be the preference relation represented by $u$. This preference relation satisfies $x\succ y \Longrightarrow x+a\succsim y+a$ for all $x,y,a\in\mathbb R$. But let $x=0$, $y=1$, and $a=-1$. Then $x\sim y$, but $y+a=0\succ -1=x+a$, thus $x+a\not\succsim y+a$ and $\succsim$ is not ...
2
This is not true.
Let $n=1$ and define $u(x)=\min{\{x,0\}}$. Let $\succsim$ be the preference relation represented by $u$. This preference relation is continuous and convex. We also have $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R$. But let $x=0$, $y=1$, and $a=-1$. Then $x\sim y$, but $y+a=0\succ -1=x+a$, thus $x+a\nsim y+a$ and $\...
1
It is not true. Let us consider $\mathbb{R}^2$ so bundles are $x = (x_1,x_2)$.
Consider the preference:
(i) If $x_1 \leq 0$, preferences are lexicographic, i.e.
$$
x \succ y \Leftrightarrow
\begin{cases}
x_1 > y_1 \\
\text{ or } \\
x_1 = y_1 \text{ and } x_2 > y_2
\end{cases}
$$
(ii) If $x_1 \geq 0$, $u(x_1,x_2)=x_1+x_2$.
Notice that no ...
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