1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $\beth_{\alpha+\omega}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.