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May 8, 2020 at 8:46 history edited YCor CC BY-SA 4.0
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Nov 20, 2009 at 16:15 comment added Mike Shulman I think a variant of your answer does work if you allow "structure" with proper-class-sized languages. Then you could define a language containing a predicate U_x for EVERY set x, and equip a set B with this structure where U_x holds of x precisely when x is in B. But while I'm happy with infinitary languages, I'd like them to be set-sized!
Nov 20, 2009 at 16:12 comment added Mike Shulman It's because of the ∃-∀ versus ∀-∃. Your answer provides, for any given set B, a type of structure such that B admits a rigid structure of that type. What I want is one type of structure, such that for every set B, B admits a rigid structure of that type.
Nov 20, 2009 at 13:38 comment added Joel David Hamkins Since you don't seem particularly interested in the binary relation case, which seems central to me, I posted Question <mathoverflow.net/questions/6262>. (I'm a newbie here, so please tell me if that isn't the right way to do things.) As you are entertaining infinitary languages, I'm not sure why my answer doesn't provide one "type" of structure in your sense. Call it, "Sets, equipped with many unary relations". In any case, I predicted you wouldn't like my answer! :-)
Nov 19, 2009 at 22:21 comment added Mike Shulman In the foundation example, we could take the type of structure to be "a well-founded extensional relation equipped with a chosen subset" and let the forgetful functor to Set pick out the subset.
Nov 19, 2009 at 22:20 comment added Mike Shulman The title of the question was misleading, but I think if you read the whole thing, you'll see that what I want is one type of structure which is rigid for all sets, i.e. ∃-∀ not ∀-∃. I'm perfectly happy with infinitary languages and even higher-order structure, and I don't care whether it's a binary relation or anything like that. I tried to make this clearer in my edit.
Nov 19, 2009 at 19:55 history answered Joel David Hamkins CC BY-SA 2.5