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Is $\aleph_0^{\aleph_0}=2^{\aleph_0}$ or $\aleph_0^{\aleph_0}>2^{\aleph_0}$ Why?

Loronegro
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  • See also this post – amWhy Feb 28 '13 at 03:38
  • Is $\aleph_0 + 1 \gt \aleph_0$ ? Nonsense, right? Why then is $2^{\aleph_0} \gt \aleph_0$ ? I could mock this subject matter. HA. I don't grasp infinite cardinalities of "different sizes". I'm inclined to argue these notions are themselves fallacies of reason. – CogitoErgoCogitoSum Feb 28 '13 at 03:47
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    @CogitoErgoCogitoSum: I don't think anyone here will argue that cardinal arithmetic is a bit wonky, and often counterintuitive, but it's hardly fallacious. If you don't understand why $2^{\aleph_0}>\aleph_0$, I recommend you read this wonderful and intuitive answer describing why a power set is always strictly larger than the starting set. P.S.: Your username cracks me up. – Cameron Buie Feb 28 '13 at 03:59
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    @Cogito: I'm sure many set theorists could mock you for not understanding, but they... ahem... have more class than that (this is a joke). – Zev Chonoles Feb 28 '13 at 04:50
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    @CogitoErgoCogitoSum Such skepticism does seem appropriate from someone whose username suggests that he/she doubts his/her very existence... – Trevor Wilson Feb 28 '13 at 05:34
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    @Zev: I hold mocking someone only after they post several anti-Cantorian and matheological questions and papers to arXiv (or viXra). :-) – Asaf Karagila Feb 28 '13 at 06:17

1 Answers1

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$\aleph_0^{\aleph_0}\le\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}\le\aleph_0^{\aleph_0}$.

Brian M. Scott
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