How can I see that $$2^{2^\lambda}>2^\lambda$$ ? Is it used here that $\lambda \geq 2^{\aleph_0}$ ? The reference is here, pages 4 and 8.
2026-03-31 07:12:23.1774941143
Simple cardinal arithmetic
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Let $X$ be a set of cardinality $2^\lambda$, then there is no bijection between $X$ and its powerset; that is, $2^{2^\lambda}=2^{|X|}>|X|=2^\lambda$ by Cantor's theorem. This holds for any cardinal $\lambda$.