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We show that for each countable ordinal , there is a operator that, for all of Scott rank at most , takes copies of to copies of , where is the tree of tuples corresponding to . For finite , we show that this is best possible. The proof that there is a operator uses Harrington’s method of workers. The sharpness result, for finite , uses recent work of Harrison-Trainor on ``unfriendly jump inversion.‘’

生成讲解博客的中文版本

生成AI智能校验的中文版本