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Joel David Hamkins's user avatar
Joel David Hamkins's user avatar
Joel David Hamkins's user avatar
Joel David Hamkins
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Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?)
Well, 2^kappa has the same kind of linear order as 2^omega, namely, the lexical order (smaller = first difference bit is smaller). I agree that the method of this answer doesn't seem to generalize much. To make a direct generalization work for 2^kappa, you would seem to need a subset equinumerous with 2^{<kappa}, to get access to the neighborhoods. But what do you do in the other cases? Is the lexical order rigid if there is no subset equinumerous with 2^{<kappa}? And even if you can handle subsets of 2^kappa for every kappa, I don't see how this handles all sets.
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Actions of finite permutation groups on hereditarily finite sets.
Also, your basic set-up presumes that the objects in M are not themselves sets, since otherwise your action is not well-defined. I guess you want to regard the objects of M as urlements for the purposes of this construction.
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Actions of finite permutation groups on hereditarily finite sets.
I think you want V_{alpha+1}(M)=V_alpha(M)\cup P(V_alpha(M)). i.e. Missing M.
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Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?)
Thanks! I admire Vopenka very much, so I'm pleased he answered this. I'll get back to you once I look this up...