König's theorem (set theory) implication

Question

How does König's theorem imply $\quad\aleph_{\omega} \neq \beth_1$?

Answer 1

Since $cof(\aleph_\omega)=\omega$, Konig's lemma tells us that $\beth_1\not=\aleph_\omega$. To see this, suppose we take the lemma in the form $$\forall i(m_i<n_i)\implies \sum_{i\in I} m_i<\prod_{i\in I} n_i,$$ and suppose $\beth_1=\aleph_\omega$. Then take $I=\omega$, $m_i=\aleph_i$, $n_i=\aleph_\omega$. The sum $\sum_{i\in I} m_i$ is just $\aleph_\omega$; meanwhile, if $\beth_1=\aleph_\omega$, the product $\prod_{i\in I}n_i$ is just $(\aleph_\omega)^\omega \color{red}= \beth_1^\omega=\beth_1$ (the red equality is our assumption for contradiction). But this contradicts Konig's lemma.

Note that it does not tell us anything else - it is consistent that $\beth_1$ is bigger, or smaller, than $\aleph_\omega$.

Surprisingly, Solovay showed that this is essentially the only restriction on $\beth_1$: if $\kappa$ has uncountable cofinality, then it is possible that $\beth_1=\kappa$. Note that this takes a lot of work to make precise, but e.g. it is consistent with ZFC that $\beth_1=\aleph_{\omega_{17}}$. (Next, Easton came along and showed much more than this - see https://en.wikipedia.org/wiki/Easton%27s_theorem).

Answer 2

Note that $\beth_1^{\aleph_0}=\beth_1$. On the other hand, $\operatorname{cf}(\aleph_\omega)=\aleph_0$, and therefore $\aleph_\omega^{\aleph_0}>\aleph_\omega$.

So if $\beth_1=\aleph_\omega$, you have both $\beth_1^{\aleph_0}=\beth_1$ and $\beth_1^{\aleph_0}\neq\beth_1$.

In fact, this argument shows that any cardinal whose cofinality is $\aleph_0$ cannot be the continuum. Solovay showed, however, that this is the only limitation on what can the continuum be. Namely, if $\kappa$ is a cardinal whose cofinality is uncountable, then it is consistent that $\beth_1=\kappa$.