MrMind comments on Open thread, Oct. 12 - Oct. 18, 2015 - Less Wrong

5 Post author: MrMind 12 October 2015 06:57AM

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Comment author: MrMind 13 October 2015 08:20:08AM 0 points [-]

Yes. It's not the Choice axiom which is problematic, but the infinity itself. So it doesn't mater if ZF or ZFC.

I doubt that any proof in FAI will use infinitary methods.

Comment author: JoshuaZ 13 October 2015 12:55:23PM 1 point [-]

I'm not sure why you think that. This may depend strongly on what you mean by an in infinitary method. Is induction infinitary? Is transfinite induction infinitary?