How can we show there is a set whose cardinality is greater than $\cal P^n(\Bbb N)$ for every natural number $n$?

Question

I haven't studied properly the theory of infinities yet.

Let $A_0$ denote the set of natural numbers. Let $A_{i+1}$ denote the set whose elements are all the subsets of $A_i$ for $i=0,...,n,...$

I understand well that the cardinality of $A_{i+1}$ is always greater than the cardinality of $A_i$ for all $i \in \mathbb{N}$.

Which is the simplest argument which proves that there exists a set whose cardinality is greater than $A_i$ for all $i \in \mathbb{N}$?

Thanks.

Answer 1

Let $X = \bigcup_{i=0}^\infty A_i$ Then $A_i\subseteq X$ for all i so $|A_i|\leq |X|$ for all i. Now, consider the powerset of $X$. Then we have $|A_i|\leq |X|<|P(X)|$.

Answer 2

Your question is closely linked to Beth numbers, their definition is:

1. $\beth_0 = \aleph_0 =$ the cardinality of the non-negative integers
2. For a successor ordinal $\alpha +1$ put $\beth_{\alpha+1} =$ the cardinality of the power-set of $\beth_{\alpha}$
3. For a limit ordinal $\delta$ put $\beth_\delta = \bigcup_{\alpha\lt\delta} \beth_\alpha$

What you're asking about is $\beth_\omega$.

Further reading: Wikipedia page on Beth number.