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Let X be a metric space and $\succeq$ be a preference relation on X. The preference relation is continuous if the sets $\succeq (y) =\{x: x \succeq y\}$ and $\preceq (y) = \{x : x \preceq y\}$ are closed for every $y$. Assume that $B \subseteq X$ is such that for any collection of closed subsets of B, $\{C_{\alpha}\}_{\alpha\in A}$ such that $\cap_{\alpha\in F}C_{\alpha}$ is not empty for any finite $F \subset A$ then $\cap_{\alpha\in A}C_{\alpha}$ is not empty. Show that if $\succeq$ is continuous, then B has a best element, i.e. $\exists$ $x^\ast\in B$ such that $x^\ast \succeq$ y for all $y \in B$.

I am finding this question very abstract and difficult to wrap my head around. Unfortunately I am not able to even start it. Any explanations of the solution would be greatly appreciated.

  • What definition of preference relation are you using? – Brian M. Scott Oct 29 '14 at 03:08
  • From the context of my notes, preference relations are defined as preferences over a consumption set X. x $\succeq$ y means the agent prefers x to y. –  Oct 29 '14 at 05:50
  • That much I knew; the question is exactly what axioms you're assuming that it satisfies. You can't hope to prove anything unless you know what assumptions are available to work with. – Brian M. Scott Oct 29 '14 at 06:41
  • Sorry. We assume the preference relation is complete and transitive. –  Oct 29 '14 at 07:18

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Here's a sketch of the argument; see if you can fill in the details.

For each $x\in B$ let $C_x=B\,\cap\succeq\!(x)$. Use the completeness of the preference relation to show that if $F$ is a finite subset of $B$, then $\bigcap_{x\in F}C_x\ne\varnothing$. Conclude that $\bigcap_{x\in B}C_x\ne\varnothing$, and show that if $b$ is any member of this intersection, then $x\preceq b$ for all $x\in B$.

Brian M. Scott
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