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

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!

• 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 $\omega$. Jun 13, 2020 at 10:09

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 $$\aleph_0<\operatorname{cf}(2^{\aleph_0})$$.