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It's pretty straightforward to say that there is an infinite number of different sizes of infinity, but then I thought, "What size of infinity is that?"

My thoughts are that the number of unique cardinalities is equivalent to the number of real numbers, based on the fact that the cardinalities can always be ordered by increasing size. I don't really know how to prove this, though. It's mostly based on intuition, which isn't very reliable when talking about uncountably infinite sets.

I originally asked a somewhat related question at a different (and not math-oriented) forum, and the users there told me that it is not possible to talk about the number of cardinalities without talking about the set of all sets, which forms a paradox. If a set were to contain all of the different sizes of infinity, it would have to contain its own power set, which isn't possible.

However, I'm not completely convinced that it is not possible to talk about a set of all of the cardinalities. Sure, a cardinality represents a size of infinity, but I think that it should be possible to have a set of the cardinalities without having the set actually contain the various infinities. Would this avoid the above paradox?

So, is it possible to measure the number of different sizes of infinity, and what would that size be?

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Also this question. –  Arthur Fischer Jan 21 '13 at 17:20
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infinitely many –  AndreasS Jan 21 '13 at 23:16

2 Answers 2

We don't really talk about "infinities", instead we talk about "cardinalities". Cardinality of the a set is the mathematical way of saying how large it is. Of course infinity could easily just mean $\infty$ which is a formal symbol representing a point larger than all real numbers (but the notion can be transferred to other contexts as well). This is not the same sort of infinity as infinite cardinalities. Infinite cardinalities are a whole other beast, and they are related to set theory (as we measure the size of sets, not the length of an interval).

Cantor's theorem tells us that given a set there is always a set whose cardinality is larger. In particular given a set, its power set has a strictly larger cardinality. This means that there is no maximal size of infinity.

But this is not enough, right? There is no maximal natural numbers either, but there is only a "small amount" of those. As the many paradoxes tell us, the collection of all sets is not a set. It is a proper class, which is a fancy (and correct) way of saying that it is a collection which is too big to be a set, but we can still decide whether or not something is in that collection.

In a similar fashion we can show that the collection of all cardinalities is not a set either. If $X$ is a set of sets, $\bigcup X=\{y\mid\exists x\in X. y\in x\}$ is also a set, and its cardinality is not smaller than that of any $x\in X$. By Cantor's theorem we have that the power set of $\bigcup X$ has an even larger cardinality.

What the above paragraph show is that given a set of cardinals, we can always find a cardinal which is not only not in that set, but also larger than all of those in that set. Therefore the collection of possible cardinalities is not a set.

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Ok, Asaf, here are some test questions to make this formal: Given a set $X$, are there $X$-many different cardinalities? If not, are there $Y$-many cardinalities for some $Y$ that surjects onto $X$? –  Andres Caicedo Jan 21 '13 at 17:46
    
This doesn't actually address why the class of all cardinalities is a proper class. It's not because there's no maximal cardinal; it's because any set of cardinals can be used to construct a larger cardinal that can't be in the set, so no set of cardinals can ever contain them all. –  jwodder Jan 21 '13 at 20:28
    
@jwodder: Yes, you are right. I saw Andres' comment but didn't have the time (or means) to edit my answer. I have now. –  Asaf Karagila Jan 21 '13 at 23:21
    
@Andres: I modified my answer to have a slightly better argument. –  Asaf Karagila Jan 21 '13 at 23:21
    
I see. There is still a minor technicality: Given a collection of cardinalities, we may not be able to pick a set of representatives, so it seems we need foundation (and replacement) to go from a set $Y$ of cardinalities, to an appropriate set $X$ whose union has cardinality larger than those in $Y$. Needing replacement is natural, but I do not know if we can circumvent foundation. –  Andres Caicedo Jan 21 '13 at 23:26

If $A$ is a set, then the power set $P(A)$ is a set of bigger cardinality. If $\{A_i\}_{i\in I}$ is a family of sets, then $P(\bigcup_{i\in I}A_i)$ is a set of bigger cardinality than any of the $A_i$. This allows us to define an infinite set $F(a)$ for each ordinal $a$ such that $a<b$ implies that $F(a)$ has smaller cardinality than $F(b)$. To do so, let

  • $F(\emptyset)=\mathbb N$,
  • $F(a)=P(F(b))$ if $a=b+1$ is the successor of $b$,
  • $F(a)=P(\bigcup_{b<a} F(b))$ if $a$ is a limit ordinal

Now if a set $S$ were able to enumerate all infinite cardinalities, this would give us an injective map from the proper class of all ordinals into this set, which is absurd.

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