What is the first cardinal which has countable cofinal and greater than $\aleph_1$?

Question

What is the first cardinal which has countable cofinal and greater than $\aleph_0$? Is it $\aleph_\omega$? I'm not sure. Thanks for any help!

Answer 1

It is $\aleph_{\omega}$. Let $\langle \alpha_n : n < \omega \rangle$ be a sequence of ordinals less than $\aleph_N$ (where $N>0$).

Since each $\alpha_n < \aleph_N$, we must have $$\left| \bigcup_{n < \omega} \alpha_n \right| \le \sum_{n < \omega} \left| \alpha_n \right| \le \sum_{n < \omega} \aleph_{N-1} = \aleph_0 \cdot \aleph_{N-1} = \aleph_{N-1} < \aleph_N$$

Thus $\aleph_N$ cannot have countable cofinality for $0<N<\omega$.

Answer 2

Well, each (infinite) cardinal below $\aleph_\omega$ is either $\aleph_0$ (which is not greater than $\aleph_1$); or is uncountable and regular (by virtue of being a successor cardinal), but then has uncountable cofinality. Since $\aleph_\omega$ has countable cofinality and is greater than $\aleph_1$, this is the least cardinal fitting the criteria.