How many different sizes of infinity are there?

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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– Amzoti Jan 21 '13 at 17:12
Also this question. – arjafi Jan 21 '13 at 17:20
infinitely many – AndreasS Jan 21 '13 at 23:16

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 $X$? – Andrés E. Caicedo Jan 21 '13 at 17:46
@user1729: Actually, there is a Gimel function too. But you can use $\daleth$... :-) – Asaf Karagila Aug 20 '14 at 10:48
@Andrey: Your comment is very unclear. What does it mean "a set of numbers"? Can you find a natural number which is not in the set of all the natural numbers? Can you find a real number which is not in the set of all the real numbers? What you can do however, is show that for every finite set of natural/rational/real/complex numbers, there is a number which does not lie in that set. Therefore, we can conclude, the set of all natural/rational/real/complex numbers is not finite. – Asaf Karagila Apr 7 '15 at 21:54
@Andrey: No, it very much does not work with infinite sets. The sum of all the natural numbers is far from a natural number (and it's not $-\frac1{12}$ either!). Sets to classes are like finite subsets of $\Bbb N$ to infinite subsets of $\Bbb N$. – Asaf Karagila Apr 9 '15 at 22:31
@Andrey: Yes, something like that. Classes are collections of sets which do not have cardinality (because, at least in $\sf ZFC$, having a cardinality means being in bijection with a cardinal which is a set). – Asaf Karagila Apr 9 '15 at 22:45

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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As the other answers have pointed out (at least within the framework of ZFC), the answer is "proper class many." Let me just point out that this isn't the end of the story.

We can ask whether two proper classes have the same size, by asking about the existence of a definable bijection, etc. In this sense the class of ordinals - or equivalently of cardinalities of well-orderable sets (=initial ordinals) - is actually small: every other proper class surjects onto it! Given a proper class $C$, consider the map $rk: C\rightarrow ON$ sending a set $x\in C$ to the unique $\alpha$ such that $x\in V_{\alpha+1}-V_\alpha$ - that is, its (von Neumann) rank. This is well-defined, and maps $C$ to a cofinal subclass $S$ of $ON$. Now we can "collapse" $S$ onto $ON$ by sending $\alpha\in S$ to $ot(\{\beta\in S: \beta<\alpha\})$; this is surjective. Composing these two maps gives a surjection from $C$ to $ON$.

Meanwhile we can construct models in which there is a proper class $C$ with no surjection from $ON$ (let alone an injection into $ON$): see Joel David Hamkins' answer to http://mathoverflow.net/questions/110799/does-zfc-prove-the-universe-is-linearly-orderable.

Finally, we could ask: can there be a proper class $C$ such that $ON$ does not inject into $C$? This is a subtler question, and class injections are weird things; but I believe the answer is yes via class forcing (I vaguely recall seeing this a long time ago, and it being relatively simple, but I can't remember the details).

The distinction between injection and surjection above might make you think, "But wait a minute! Don't we have the axiom of choice to simplify things?" Indeed we do, but the axiom of choice treats only sets, and at the level of classes "injects into" and "is surjected onto" are still distinct, a priori. We can have an "axiom of choice for classes" (called global choice), if we enlarge our language a bit to talk about classes, and this axiom implies that all proper classes have the same size.

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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 Aug 20 '14 at 8:13
@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 Aug 20 '14 at 9:34
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 Aug 20 '14 at 9:35
@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 Aug 20 '14 at 9:38
@mistermarko I don't know abut Kant, but Bell is certainly not wrong. But he never stated that it is consistent with ZF that there is "only one size of infinity". Because he has a clue about set theory. – Michael Greinecker Aug 23 '14 at 17:56

The different sizes of infinities are called transfinities. They can be captured as mathematical sets via formalized definitions. The cardinality (number of) transfinities is not itself a transfinity, it is infinity. It cannot be captured as a set. It cannot be formalized by a mathematical definition. In that sense it is truly infinity which just means an unknowable, indefinable concept.

Borrowing ideas from complexity theory we see that transfinities are compressible whereas infinity is not. Pure randomness is not compressible. The cardinality of transfinities sets is actually pure randomness which is maximum entropy. That's an interesting way to tie in entropy to the concepts of transfinities and their cardinality (number of yransfinities).

Furthermore, mathematically nothingness can be defined as the state of maximum entropy. To get philosophical (and also mathematical) the cardinality of transfinities is both the alpha and the omega.

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-1: not only does this not address the question, it's not remotely correct. No, they're not called transfinities, at least not in set theory - in fact I've never seen the term. Googling for it reveals no scholarly mathematical works on the first page of hits, and the only scholarly works I can find are on "neurophilosophy". As to the content: no, infinite cardinalities need not be definable. The connection to complexity theory makes no sense, and neither does the mention of entropy. To get philosophical, if you're going to get philosophical at least make sure it's actually relevant. – Noah Schweber May 24 '15 at 23:00