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Consider the category Top of topological spaces. Consider the contravariant functor from Top to Set sending a topological space X to the set of all opens of X. Is this functor representable?

What if you replace "open" by "closed"?

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    Of course that's the one I meant! :)2011-10-04

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Yes, it is (assuming the contravariant functor is what you mean, with the functoriality given by pull-back). Consider the topological space $T = \{0, 1\}$ with the open sets being $\{0\}$ and $T$. Then to give an open set in a topological space $X$ is the same as giving a map $X \to T$ (which sends said open set into ${0}$). The case for closed sets is the same.