Take the following definition of continuity.

The preference relation $\succsim$ over the space of lotteries $\mathcal L$ is continuous if for any $L,L',L''\in\mathcal L$, the sets $$S_1=\{\alpha\in[0,1]:\alpha L+(1-\alpha)L'\succsim L''\}$$ and $$S_2=\{\alpha\in[0,1]:L''\succsim \alpha L+(1-\alpha)L'\}$$ are both closed.

Is it necessarily true that $S_1\cup S_2=[0,1]$? If so, why?


1 Answer 1


It is.
Prior to continuity, which is a property of the preference relation, the preference relation $\succsim$ itself has been defined to be a binary relation that is characterized by transitivity, and, to begin with, by completeness.
Then if $S_1\cup S_2 \neq [0,1]$, it means that there exist some values of $\alpha$ somewhere in $[0,1]$, call them $\tilde \alpha$ for which


$$\{\tilde \alpha L+(1-\tilde \alpha)L'\succsim L''\}$$


$$\{L''\succsim \tilde \alpha L+(1-\tilde \alpha)L'\}$$

In words, for these $\tilde \alpha$'s, the pair cannot be ordered at all. But this contradicts the completeness foundation that is needed to even obtain a preference relation (as of course used in our theory. Psychologists I guess would disagree).

Also, note that completeness is defined over all conceivable pairs, even if, in a specific situation, we chose to restrict the space of lotteries to something smaller. Whether the lotteries under consideration belong to the specified lottery space, is really irrelevant. The person having the preferences has to be able to order them in any case, even as a "hypothetical" scenario (although strictly speaking, for a specific problem we have the "luxury" to impose completeness only as regards the lotteries available, while "remaining agnostic" as regards completeness if we expand the lottery space. Still this "weakening" on the imposition of the completeness axiom, does not really bring any gain).


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