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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
infinitely many –  AndreasS Jan 21 '13 at 23:16

3 Answers 3

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
@user1729: Actually, there is a Gimel function too. But you can use $\daleth$... :-) –  Asaf Karagila yesterday

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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'Kant is one of those who think that... Every finite set of F's excludes at least one F (1), though it contradicts the statement that there are only finitely many F's, is nevertheless weaker than... There is an infinite number of F's (2)'. The Age and Size of the World, Jonathan Bennett; quoted - p128, The Continuous and the Infinitesimal, John Bell. Bell then says 'Bennett implies that Kant is simply mistaken here, that in fact (1) and (2) are equivalent. But is this right? Let us bring to bear some contemporary mathematical ideas on the matter.' He then goes on to show that it is possible for a set to be 'Kantian', that is potentially but not actually infinite. He concludes '...the existence of Kantian sets is consistent with the axioms of set theory (and classical logic) as long as the axiom of choice is not assumed.' (Italics in original).

@Asaf Karagila This is verging on copyright infringement, but anyway: 'For suppose that we are given a potentially infinite set A, and we attempt to show that it is actually infinite by arguing as follows. We start by picking a member a0 of A; since A is potentially infinite, there must be a member of A different from a0; pick such a member and call it a1. Now again by the fact that A is potentially infinite, there is a member of A different from a0, a1 - pick such and call it a2. In this way we generate a list a0, a1, a2, ... of distinct members of A; so, we are tempted to conclude, A is actually infinite. But clearly the cogency of this argument hinges on our presumed ability to "pick", for each N, an element of A distinct from a0, ..., aN - an ability enshrined in the set-theoretic principle known as the axiom of choice. Now the axiom of choice is ... a perfectly consistent mathematical assumption. But ... its denial is equally consistent. In fact, it can be denied in such a way as to prevent the argument just presented from going through, that is, to allow the presence of potentially infinite sets which are not at the same time actually infinite.'

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Uhh, what?${}{}{}$ –  Asaf Karagila yesterday
@Asaf I think the point of this is a poking of your argument in the second last paragraph. But I think mistermarko needs to make his criticism clearer...(certainly, it took me a few reads to get this to parse correctly - the second paragraph is a quote from Bell's book.) –  user1729 yesterday
mistermarko, I am not notified about edits. So the next time you edit something, please ping the person to whom the edit is direction (by using @Asaf in my case, for example), and let them know of the changes. Also, and perhaps more importantly, line breaks are not a bad idea when writing a block of text. –  Asaf Karagila yesterday
@user1729: Okay, so reading this thing again, I see that the answer is that it is consistent to have infinite Dedekind-finite sets (meaning an infinite set without a countably infinite subset). What does this have to do with the question, though? –  Asaf Karagila yesterday
@Asaf I think that it is a criticism of your argument. But I can only guess how mistermarko wants the ideas to mesh. I think mistermarko might be latching on to the phrase "...and its cardinality is not smaller than that of any..." which could, perhaps, maybe, be related to saying that a "finite set of cardinalities excludes at lease one cardinality". –  user1729 yesterday

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