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I wondered whether there was any connection between The Continuum Hypothesis and The Axiom of Choice. I was not surprised to learn that they are independent. See this previous question with a similar but not identical title.

The Continuum Hypothesis & The Axiom of Choice

Reading on, I was surprised to find that the Generalized Continuum Hypothesis does imply the Axiom of Choice. Now it is less surprising that The Continuum Hypothesis does not imply the Axiom of Choice since it only deals with "small" cardinalities. This led me to the question of this post: is there a connection between The Continuum Hypothesis and The Axiom of Countable Choice?

ZF is assumed throughout.

badjohn
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1 Answers1

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The disanalogy becomes more obvious if we rephrase choice in terms of well-orderability. $\mathsf{GCH}$ implies all sets are well-orderable, but $\mathsf{CH}$ only makes sets of size $\beth_1$ well-orderable. But we prove a set of nonempty sets has a choice function by well-ordering its union. There's no rule connecting the original set being countable to its union's size. So while $\mathsf{ZF}+\mathsf{GCH}$ implies $\mathsf{AC}$, the same logic doesn't tell us $\mathsf{ZF}+\mathsf{CH}$ implies $\mathsf{AC}_\omega$ (in fact, it doesn't).

J.G.
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