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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?

Asaf Karagila
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PhiNotPi
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4 Answers4

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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.

Asaf Karagila
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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
  • 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. – Andrés E. Caicedo Jan 21 '13 at 23:26
  • (But independently of this, think about my two questions, there may be something interesting there. I do not know the answer even in very concrete cases.) – Andrés E. Caicedo Jan 21 '13 at 23:27
  • @Andres: I did write that at first, but then I realized exactly that, so I changed it to the current wording. Given a set of Scott cardinals, just take all the sets in those cardinals. It's still a set due to replacement and union. As my usual motto goes "If you can't choose one, just take everything!". As for the question without foundation, cardinals aren't really that well-defined (internally, that is) when both choice and foundations are missing, so we're stumped. – Asaf Karagila Jan 21 '13 at 23:28
  • Sure, it may be we cannot really say anymore in such a handicapped setting. Anyway, if you have any thoughts on the two questions, let me know. – Andrés E. Caicedo Jan 21 '13 at 23:31
  • Just to be clear: the collection of all cardinalities does not form a set, so they themselves have no cardinality? So there is an "undefined" number of cardinalities?... – user1729 Aug 20 '14 at 09:26
  • @user1729: The collection is not a set, just sets have cardinalities. It's like saying, the collection of all finite sets of natural numbers is not a finite set. So its cardinality is not a natural number, is it undefined? In a context where no infinite sets exist, yes, it is undefined (and also this set is a proper class now). In other contexts the answer is different. If you consider a class set theory (e.g. NBG or MK) then you can talk about cardinalities of proper classes, in a somewhat awkward way, but still. Then you can say that classes have some sort of cardinality assigned to them. – Asaf Karagila Aug 20 '14 at 09:28
  • @Asaf Ah, okay. So to take your analogue about finite sets and finite cardinals a bit further, we sort of get a third-tier of cardinals. We have the natural numbers, then the alephs, and then the class of all alephs has a cardinality, $\beth_0$ say, which gives us a third tier. Is this sort of, maybe, informally, what is going on? That we can define the cardinality of a class, if we wish, and it sits above the usual infinite cardinals just like they sit above the finite cardinals? – user1729 Aug 20 '14 at 09:52
  • @user1729: $\beth_0$ is actually $\aleph_0$. But yes, that's the idea. You have a third-tier of size. But without having means of talking about classes properly this is the same as having "infinite size" as a notion, which you may or may not be able to extend a little bit (depending on the axioms you have at your disposal). – Asaf Karagila Aug 20 '14 at 10:04
  • @Asaf Sure, okay. As $\gimel$ apparently renders, I'll call the third tier "the gimles"...so I mean "...and then the class of all alephs has a cardinality, $\gimel_0$ say,..." – user1729 Aug 20 '14 at 10:06
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    @user1729: Actually, there is a Gimel function too. But you can use $\daleth$... :-) – Asaf Karagila Aug 20 '14 at 10:48
  • Can the reasoning of your last paragraph apply to numbers? "Given a set of numbers, we can always find a number which is not only not in that set, but also larger than all of those in that set. Therefore the collection of possible numbers is not a set." – Andrey Fedorov Apr 07 '15 at 21:28
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    @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 07 '15 at 21:54
  • Sure, for natural numbers: the sum of a set of natural numbers is surely a natural number, just as the union of sets is a set. I guess that doesn't work for infinite sets of numbers, does it?

    To clarify / summarize, the proof is that if the set of cardinalities $X$ existed, $\mathcal{P}(\bigcup X)$ would have to be in there, and that's impossible.

    – Andrey Fedorov Apr 09 '15 at 22:28
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    @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 09 '15 at 22:31
  • Thanks! Could you expound on that last point? Finite subsets of $\Bbb N$ have a cardinality as do infinite subsets, although the latter is not measurable with an element in $\Bbb N$. Are classes like sets which have no cardinality? Is there any alternative measure of the size of a class, or are they all just "too big to measure"? e.g. can one set up a bijection between two classes? – Andrey Fedorov Apr 09 '15 at 22:42
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    @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 09 '15 at 22:45
  • In the last paragraph, did you mean paragraphs show or paragraph shows? – J. W. Tanner Jun 16 '23 at 13:03
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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 https://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.

Noah Schweber
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In the video below I present a challenge to the idea, currently defended by mathematicians, that there are different sizes of mathematical infinities. In addition to showing a new interpretation to Cantor's Diagonal Argument, I also show that a one-to-one correspondence between the set of natural numbers and any other infinite set is a logical possibility. The video is only 19min long and is easy to understand.

https://www.youtube.com/watch?v=vGCDdyCAMTc

video name: The biggest mistake in the history of mathematics - on how to interpret Cantor's Diagonal Argument

Thanks

  • An answer should provide an answer authored by the answerer, and the use of a link should only be used to provide more details. Here, your words depend entirely on the video you linked. link-only answers do not count as an answer. – amWhy Feb 15 '22 at 20:37
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