Cofinality of $2^{\aleph_\omega}$

Question

Is the following statement correct: $\operatorname{cf} (2^{\aleph_\omega})=\aleph_0$? It appears in the book "Introduction To Set Theory" by Thomas Jech (page 165).

Answer 1

Assuming the Axiom of Choice, $\operatorname{cf}(2^{\aleph_\omega})\gt\aleph_0$. This is a consequence of König's theorem, which says that, if $m_i\lt n_i$ holds for each $i\in I$, then $$\sum_{i\in I}m_i\lt\prod_{i\in I}n_i.$$ Hence, if $m_i\lt2^{\aleph_\omega}$ for each $i\lt\omega$, then $$\sum_{i\lt\omega}m_i\lt(2^{\aleph_\omega})^{\aleph_0}=2^{\aleph_\omega\cdot\aleph_0}=2^{\aleph_\omega}.$$ A similar argument shows that $\operatorname{cf}(2^{\aleph_\omega})\gt\aleph_\omega.$

I don't have a copy of Jech's Introduction to Set Theory, but if it really says $\operatorname{cf}(2^{\aleph_\omega})=\aleph_0$ that's obviously a typo for $\operatorname{cf}(\aleph_\omega)=\aleph_0$. We know that $2^{\aleph_0}\ne\aleph_\omega$ because $\operatorname{cf}(2^{\aleph_0})\gt\aleph_0=\operatorname{cf}(\aleph_\omega)$.