# cardinality of a countable sequence of powersets

Given a set $S_1$ of cardinality $\kappa$, we can construct the sequence $\langle S_1, S_2, S_3 ... \rangle$, where $S_i = \wp(S_{i-1})$, for all $i > 1$. If $S$ is finite, so that $\kappa < \aleph_0$, then I take it that $$\Big| \bigcup_{i = 1}^\infty S_i \Big| = \aleph_0$$ What if $\kappa = \aleph_0$? What is the cardinality of $\bigcup_{i=1}^\infty S_i$ then? What if $\kappa = \beth_1$? Are there any general definitions or theorems about the cardinality of sets like these?

## 1 Answer

It’s nicer to start at $0$. If $|S_0|=\aleph_0=\beth_0$, then $|S_1|=2^{\beth_0}=\beth_1$, and in general $|S_n|=\beth_n$. Thus,

$$\left|\bigcup_{n\in\omega}S_n\right|=\beth_\omega\;.$$

In general, if $|S_0|=\beth_\alpha$, then the union will have cardinality $\beth_{\alpha+\omega}$. If $|S_0|$ is not a beth number, it gets a bit messier, but not much. If $\beth_\alpha\le\kappa<\beth_{\alpha+1}$, then $\beth_{\alpha+1}\le 2^\kappa\le\beth_{\alpha+2}$, and by induction $\beth_{\alpha+n}\le|S_n|\le\beth_{\alpha+n+1}$ for each $n\in\omega$. Thus, the union will have cardinality $\beth_{\alpha+\omega}$.