What is $\aleph_1^{\aleph_0}$?
Can anyone shed some light. I'm not sure if this is provable or even has a value. Thanks
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Since $\aleph_1\leq 2^{\aleph_0}$, it follows that $$2^{\aleph_0} \leq \aleph_1^{\aleph_0} \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0},$$ and hence we conclude that $\aleph_1^{\aleph_0} = 2^{\aleph_0}$.
More generally, for any infinite cardinal $\kappa$, $2^{\kappa}=\lambda^{\kappa}$ for all cardinals $\lambda$ with $2\leq \lambda\leq \kappa^+$. See here. Since $\aleph_1=(\aleph_0)^+$, the result also follows from that observation.
Note that some of the assertions above may involve or require the Axiom of Choice.
Arturo Magidin
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2+1. To elaborate for the OP on the last line of this answer: while $\mathsf{ZF}$ alone proves the existence of a surjection from $2^{\aleph_0}$ to $(\aleph_1)^{\aleph_0}$ it does not prove the existence of an injection going the other way - or even an injection from $\aleph_1$ to $2^{\aleph_0}$. $\mathsf{ZF}$ can prove that each of $2^{\aleph_0}$ and $(\aleph_1)^{\aleph_0}$ surjects onto the other, but without choice we can't prove that "surjections both ways implies bijection" (while injections both ways are enough). – Noah Schweber Jun 18 '20 at 20:39
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@NoahSchweber: In fact, “surjections both ways implies bijection” implies the Axiom of Countable Choice. See this MO post and references therein – Arturo Magidin Jun 18 '20 at 20:44
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1Of course, I just wanted to keep things snappy. (Until the OP's comfortable with AC on its own, bringing in its fragments is likely to just get messy.) – Noah Schweber Jun 18 '20 at 20:52
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1@Arturo I don’t see that result in the linked. All I see demonstrated in the references is that it implies every infinite set has a countable subset. – spaceisdarkgreen Jun 18 '20 at 21:23
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@spaceisdarkgreen: ah, you are correct; Asaf corrected that impression in the comments, noting it is weaker than countable choice. Thank you. – Arturo Magidin Jun 18 '20 at 21:25