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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.

2 Answers 2

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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 $|A_i|\leq |X|<|P(X)|$.

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    Cardinals need not be alephs. But AC implies "Every cardinal is an aleph".2011-05-15
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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.

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    @Theo: Yes, I was thinking about. To be fair, there is this one edit I was really trying to avoid. Oh well... :P2011-08-15