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Let $A_0$ be the empty set and $A_n := \mathcal{P}(A_{n-1})$ for $n \in \mathbb{N}$. I have to determine $A_n$ and $|A_n|$.

Using the definition of the power set I get \begin{align} A_1 & = \left\{ \emptyset \right\} & |A_1| = 1 \\ A_2 & = \left\{ \emptyset, \left\{ \emptyset \right\} \right\} & |A_2| = 2 \\ A_3 & = \left\{ \emptyset, \left\{ \emptyset \right\}, \left\{ \left\{ \emptyset \right\} \right\}, \left\{ \emptyset, \left\{ \emptyset \right\} \right\} \right\} & |A_3| = 4 \end{align}

and so on. Since $|\mathcal{P}(A)| = 2^{|A|}$, the cardinality of $|A_n|$ is determined by the sequence \begin{align} |A_0| & = 0 \\ |A_n| & = 2^{|A_{n-1}|} \quad \forall n \in \mathbb{N} \end{align} So with a (clumsy) dot notation, $|A_n| = 2^{2^{2^{\vdots^{2^0}}}}$.

However, I did not find any explicit form of either sequences (which is somewhat suggested by the formulation "determine $A_n$ and $|A_n|$").

Thanks in advance for any help!

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    $\begingroup$ Your answer is entirely correct. You've determined $|A_n|$; it's just an accident of mathematical history that there isn't a neater notation for the answer. The same is true for $A_n$. $\endgroup$ – Michael Weiss Mar 13 '14 at 8:33
  • $\begingroup$ @MichaelWeiss thanks, that was what I suspected. $\endgroup$ – Gaste Mar 13 '14 at 8:55
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As Michael Weiss noted in the comments:

Your answer is entirely correct. You've determined $|A_n|$; it's just an accident of mathematical history that there isn't a neater notation for the answer. The same is true for $A_n$.

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