There is a similar question in this site but I am not satisfied with the answer, which is basically the same as the proof in the mentioned textbook.

The book(Karel Hrbacek&Thomas Jech, Introduction to Set Theory 3e, p165) states a lemma: For every $\alpha$, $\text{cf}(2^{\aleph_\alpha})>\aleph_\alpha$. Then it asserts that $2^{\aleph_0}$ cannot be $\aleph_\omega$, since $\text{cf}(2^{\aleph_\omega})=\aleph_0$. But I can't see the connection. According to the lemma, $\text{cf}(2^{\aleph_\omega})$ should be larger than $\aleph_\omega>\aleph_0$, how can it equal $\aleph_0$?

On the other hand, I can't see why $\text{cf}(2^{\aleph_\omega})=\aleph_0$ is false either. Since $2^{\aleph_\omega}=\lim\limits_{n\rightarrow\omega}2^{\aleph_n}$, it is the limit of an increasing sequence of ordinals of length $\omega$, so its cofinality should not be greater than $\aleph_0$. Is there something wrong within this reasoning?

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    $\begingroup$ I don't have that book, but the assertion you quoted must have a typo. I'm sure they meant to sey that $2^{\aleph_0}$ cannot be $\aleph_\omega$ since $\operatorname{cf}(\aleph_\omega)=\aleph_0$, while (by the lemma) $\operatorname{cf}(2^{\aleph_\omega})\gt\aleph_\omega.$ $\endgroup$ – bof Jun 9 '15 at 23:51

All you can say about $\operatorname{cf}(2^{\aleph_\omega})$ is that it's some regular cardinal $\kappa$ such that $$\aleph_{\omega+1}\le\kappa\le2^{\aleph_\omega}.$$ I think you need the axiom of choice to say even that much.

What's wrong with your reasoning is the unwarranted assumption that $$2^{\aleph_\omega}=\lim_{n\to\omega}2^{\aleph_n}.$$ Ordinal exponentiation is continuous, but cardinal exponentiation is not; e.g., $$2^{\aleph_0}\ne\lim_{n\to\omega}2^n.$$

In fact, there are models of set theory (with choice) in which the equality $$2^{\aleph_\omega}=\lim_{n\to\omega}2^{\aleph_n}$$ holds, but in that case we have $$2^{\aleph_k}=2^{\aleph_{k+1}}=2^{\aleph_{k+2}}=\cdots=2^{\aleph_\omega}$$ for some $k\lt\omega,$ i.e., the sequence $\{2^{\aleph_n}\}_{n\lt\omega}$ is not a strictly increasing sequence, but instead is eventually constant.


As bof says, the statement "$2^{\aleph_\omega}=\lim 2^{\aleph_n}$" is false - to see why, consider the following analogous claim: $$2^\omega=\lim 2^n.$$ They are each wrong for precisely the same reason.

EDIT: Specifically, if we try to "cover" $2^\omega$ by $\bigcup 2^n$ in the obvious way, we miss all the infinite subsets of $\omega$. Similarly, if we try to cover $2^{\aleph_\omega}$ by $\bigcup 2^{\aleph_n}$ we miss all the cofinal subsets of $\aleph_\omega$. Yes, each cofinal $X\subseteq \aleph_\omega$ can be written as $X=\bigcup Y_i$ with each $Y_i\subseteq \aleph_i$, but this doesn't let us build the bijection you might want, for the same reason that saying "each subset of $\omega$ is a limit of finite sets" doesn't let us argue that $2^{\aleph_0}$ is countable.


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