Given infinite cardinal numbers $\mathfrak{m<n}$, is it true that $2^\mathfrak{m}<2^\mathfrak{n}$?
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6That's independent of ZFC – Alessandro Codenotti Dec 20 '19 at 14:30
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2This is definitely a dupe: I've seen Asaf answer this several times. – Rushabh Mehta Dec 20 '19 at 14:35
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2As per @AlessandroCodenotti's comment, see the fourth bullet point here (sadly, with no reference). See also this answer. For a proof, see Jech 2003. – J.G. Dec 20 '19 at 14:36