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Cantor’s theorem states that $|\mathcal{P}(A)| > |A|$ for any set $A$.

As a special case of this, we have $|\mathcal{P}(\mathbb{N})| > |\mathbb{N}|$. If we denote the power set of the naturals by $\mathcal{P_1}$, then $|\mathcal{P}(\mathcal{P_1})| > |\mathcal{P_1}|$. Now denote the power set of $\mathcal{P_1}$ by $\mathcal{P_2}$. Applying the theorem again, we have $|\mathcal{P}(\mathcal{P_1})| > |\mathcal{P_2}|$. We can repeat this process forever, with each power set, $\mathcal{P_n}$, having strictly greater cardinality than the last, $\mathcal{P_{n-1}}$.

Clearly this shows that there is an infinite number of different cardinalities of infinite sets (an infinity of infinities). My first thought is then to wonder about the nature of this infinite quantity.

I have two questions:

  1. What is the cardinality of the set of different cardinalities of infinite sets? In other words, what kind of infinity is the number of kinds of infinity? It seems like it would be countable infinity to me, as we clearly have a bijection between the naturals and these cardinalities. However, my question below may cause a problem with this argument.
  2. In performing this iterative process on $\mathbb{N}$, are we actually covering all possible kinds of infinity? Can we be sure that there don’t exist any kinds of infinity that won’t at some point be generated by this process?

A side question regarding the definition of ‘uncountable’: Does uncountable infinity refer to any infinity that isn’t $|\mathbb{N}|$, or is it strictly $|\mathbb{R}|$?

I hope this question makes sense and that I’m not asking something pointless or unanswerable.

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marked as duplicate by Asaf Karagila set-theory Apr 20 at 0:32

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  • $\begingroup$ Regarding your question about the definition of "uncountable": I believe it refers to any infinity that isn't $\mathbb{N}$? I actually was confused about the same thing for years, and I thought that it means the cardinality of $\mathbb{R}$. But now I'm $99\%$ sure it's otherwise. $\endgroup$ – Ovi Apr 20 at 0:24
  • $\begingroup$ Uncountable means infinite but not $|\mathbb{N}|$. $\endgroup$ – Morgan Rodgers Apr 20 at 0:29
  • $\begingroup$ Isn't this undecidable in ZFC? $\endgroup$ – Brian Tung Apr 20 at 0:30
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    $\begingroup$ Surely there are more, and perhaps better, duplicates. I leave you with the exercise of searching for them. Those three can make quite a good stepping stone for that. $\endgroup$ – Asaf Karagila Apr 20 at 0:33
  • $\begingroup$ Of course I had a look for duplicates before posting, but I couldn’t find any. I’ll search further. $\endgroup$ – 雨が好きな人 Apr 20 at 0:36

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