Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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

share|improve this answer
    
I just wanted to add that this construction is nice because for ANY collection (possibly uncountable) of sets (so long as the index set is, in fact, a set), one can form the corresponding set "X" by the axiom of replacement and the axiom of union. Thus, one can always find a bigger set than all the sets in any particular collection. The downside is that this construction uses the axiom of choice (in asserting P(X) has a cardinality). There ARE ways I know of creating larger and larger cardinalities without choice, but they are all quite messier than this. –  Jason DeVito Aug 22 '10 at 21:40
    
Your proof works fine without choice. You don't need to explicitly mention cardinalities (you can just talk about sets and injections directly), and you don't need choice for cardinalities anyway (that's just the simplest way of doing them). –  Chris Eagle May 15 '11 at 7:05
    
Cardinals need not be alephs. But AC implies "Every cardinal is an aleph". –  GEdgar May 15 '11 at 12:32

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.

share|improve this answer
    
Since \beth is now available and this question got bumped by someone for some reason, I made the plain text into LaTeX and took the liberty of introducing some words. –  t.b. May 15 '11 at 10:40
    
@Theo: Many thanks :-) –  Asaf Karagila May 15 '11 at 12:14
    
Heh! I just earned the excavator badge because of this edit. This badge-business is getting sillier and sillier :) Wanna have a badge? Go edit a very old inactive post... –  t.b. Aug 15 '11 at 9:53
    
@Theo: Yes, I was thinking about. To be fair, there is this one edit I was really trying to avoid. Oh well... :P –  Asaf Karagila Aug 15 '11 at 9:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.