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Let $2^{\aleph_0} > \aleph_\omega$. Then $\aleph_\omega^{\aleph_0} = 2^{\aleph_0}$.

My attempt seems flawed to me. Could someone please clarify?

I attempt $\aleph_\omega^{\aleph_0} \leq 2^{\aleph_0}$:

Let $\aleph_\omega^{\aleph_0} \leq \aleph_\omega^{\aleph_\omega} = 2^{2^{\aleph_0}} = 2^{2\cdot\aleph_0} = 2^{\aleph_0}$.

The flaw seems to be the second equality $2^{2^{\aleph_0}} = 2^{2\cdot\aleph_0}$.

Maybe on a more general note: How do you attack problems like the one above? I do not have much working experience with cardinals and was unsure whether to build up some kind of bijection (which seemed pretty complicated) or try to go at it using arithmetic rules for cardinals.

Cheers

lowcard
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    $2\leq \aleph_{\omega} \lt 2^{\aleph_0}$, so $2^{\aleph_0}\leq \aleph_{\omega}^{\aleph_0} \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0}$. – Arturo Magidin Oct 18 '22 at 18:07
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    What you claim is not just a flaw, it is definitely wrong. By Cantor's Theorem, $2^{\aleph_0} \lt 2^{2^{\aleph_0}}$, so you cannot have $2^{2^{\aleph_0}} = 2^{2\aleph_0} = 2^{\aleph_0}$. – Arturo Magidin Oct 18 '22 at 18:09
  • I see, thanks for clearing that up. I could not see the Cantor anymore in all those alephs. – lowcard Oct 18 '22 at 19:52
  • @ArturoMagidin Could you elaborate on why the strict inequality $\aleph_\omega < 2^{\aleph_0}$ implies a $\leq$ inequality? Is there a typical case where we can obtain an equality? – lowcard Oct 18 '22 at 19:56
  • In general, cardinal exponentiation will not respect equalities. For example, $2\lt \aleph_0$, but $2^{\aleph_0}=\aleph_0^{\aleph_0}$. So even if you assume $\aleph_{\omega}\lt 2^{\aleph_0}$, you can only conclude $\aleph_{\omega}^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}$. – Arturo Magidin Oct 18 '22 at 20:08
  • In general, for infinite cardinal $\kappa$, we get that $2^{\kappa}=\lambda^{\kappa}$ for all $\lambda$ that satisfies $2\leq \lambda\leq \kappa^+$, where $\kappa^+$ is the smallest cardinal strictly larger than $\kappa$. So that's a lot of equalities. The argument is exactly the same as the one I gave above. See here. – Arturo Magidin Oct 18 '22 at 20:11
  • About the "equality" $2^{2^{\aleph_0}} = 2^{2\cdot \aleph_0}$: You're probably thinking about the fact that $(\kappa^{\lambda})^\mu = \kappa^{\lambda\cdot \mu}$. But exponentiation is not associative! Usually $\kappa^{\lambda^\mu}$ means $\kappa^{\left(\lambda^\mu\right)}\neq (\kappa^{\lambda})^\mu$. – Alex Kruckman Oct 18 '22 at 20:54

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