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Show that $2^{\aleph_0}$ $\neq$ $\aleph_{\alpha+\omega}$ for any ordinal $\alpha$.

What I did was the following:

I first used ordinal addition where $\alpha+\omega$ = $\sup\{\alpha+n:n \in \omega\}=\sup \: \omega$ = $\omega$. Thus, $\aleph_{\alpha+\omega}$ = $\aleph_{\omega}$. Then there is a corollary in my text book that says: $\aleph_0 \in cf(2^{\aleph_0})$, so $2^{\aleph_0}$ $\neq$ $\aleph_{\omega}$, as $cf(\aleph_\omega)=\aleph_0$. Hence, $2^{\aleph_0}$ $\neq$ $\aleph_{\alpha+\omega}$.

Have I done something wrong here, as I got zero points on this task. Thanks for your help!

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  • $\begingroup$ One of the steps you do is to derive $\alpha+\omega=\omega$ for arbitrary $\alpha$. This is obviously not true since $\omega+\omega=\omega\cdot 2\ne\omega$, and $\omega_1+\omega$ isn't even in bijection to $\omega$. $\endgroup$
    – celtschk
    Jun 13, 2020 at 10:09

1 Answer 1

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You got zero points, as you should have.

It is true that $\alpha+\omega=\sup\{\alpha+n\mid n<\omega\}$ the rest is absolutely false. Note, for example, that $\omega_1+n$ is uncountable, for any $n<\omega$, but you are claiming that $\sup\{\omega_1+n\mid n<\omega\}$ is a countable ordinal. How is that even possible?

What is true, however, is that the cofinality of $\aleph_{\alpha+\omega}$ is countable, as witnessed by $\aleph_{\alpha+n}$ for $n<\omega$, being a cofinal sequence. Then we can apply the theorem stating that $\aleph_0<\operatorname{cf}(2^{\aleph_0})$.

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