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

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marked as duplicate by Namaste, Cameron Buie, Asaf Karagila, Andrés E. Caicedo, Micah Feb 28 '13 at 4:21

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  • $\begingroup$ See also this post $\endgroup$ – Namaste Feb 28 '13 at 3:38
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    $\begingroup$ @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. $\endgroup$ – Cameron Buie Feb 28 '13 at 3:59
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    $\begingroup$ @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). $\endgroup$ – Zev Chonoles Feb 28 '13 at 4:50
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    $\begingroup$ @CogitoErgoCogitoSum Such skepticism does seem appropriate from someone whose username suggests that he/she doubts his/her very existence... $\endgroup$ – Trevor Wilson Feb 28 '13 at 5:34
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    $\begingroup$ @Zev: I hold mocking someone only after they post several anti-Cantorian and matheological questions and papers to arXiv (or viXra). :-) $\endgroup$ – Asaf Karagila Feb 28 '13 at 6:17
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$\aleph_0^{\aleph_0}\le\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}\le\aleph_0^{\aleph_0}$.

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