I know that ZF set theory without the axiom of choice can prove the cardinal inequality $2^{\aleph_0} \geq \aleph_1$. That raises the question, can ZF set theory without the axiom of choice prove the generalized cardinal inequality $2^{\aleph_\alpha} \geq \aleph_{\alpha + 1}$, for all ordinals $\alpha$?
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It might be irrelevat, but by $\kappa\ge \lambda$ you mean that $\lambda$ injects into $\kappa$, right? Or do you mean the weaker assertion that $\kappa$ surjects onto $\lambda$? – Aug 17 '21 at 21:42
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2It depends how we interpret "$\ge$." It's easy to show in $\mathsf{ZF}$ that there is a surjection from $2^{\aleph_\alpha}$ to $\aleph_{\alpha+1}$ (think about well-orderings of $\aleph_\alpha$ itself); on the other hand, it's consistent with $\mathsf{ZF}$ that there is no injection $\aleph_1\rightarrow 2^{\aleph_0}$. (I've given this as a comment as opposed to an answer since I'm pretty sure this has been asked before.) – Noah Schweber Aug 17 '21 at 21:43
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1@Noah: You might as well say "that depends on how you interpret $\aleph_1$". The symbol $\leq$, and consequently $\geq$, has one meaning in the context of cardinals. Period. – Asaf Karagila Aug 17 '21 at 21:47