# What is the cardinality of the set of infinite cardinalities?

I am currently aware of only two infinite cardinalities:

• $\aleph_0 = |\Bbb N|$
• $\aleph_1 = |\Bbb R|$

Questions:

1. Is there an infinite number of infinite cardinalities?
2. If yes, is this set of cardinalities countable or uncountable?

I know that $\aleph_0<\aleph_1$ and I would tend to guess that the set of infinite cardinalities is countable at most (or possibly even finite), since there is no known element $K$ such that $\aleph_0<K<\aleph_1$.

BTW, are there any other acceptable operations (such as $+,-,\times,/$) between elements in this set?

Thanks

• $|\mathbb{R}|= \aleph_1$ is an undecidable assertion, and by definition there is no $\kappa$ such that $\aleph_0<\kappa<\aleph_1$ – jcg Mar 27 '14 at 11:09
• 1. Yes, by Cantor's theorem. 2. No. Class of all cardinals is not even a set! – Hanul Jeon Mar 27 '14 at 11:10
• 1. Because $\mathcal{P}(X)$ is always larger than $X$, you can easily get a countable number of cardinalities just by defining $X_0=\mathbb{N}$ and $X_i=\mathcal{P}(X_{i-1}$. – JiK Mar 27 '14 at 11:10
• @Julien Godawatta : Thanks. By definition? I thought that this was an unproved hypothesis... – barak manos Mar 27 '14 at 11:21
• @barakmanos Because the Cantor's paradox, this class is not a set. – Hanul Jeon Mar 27 '14 at 11:24

Firstly, kudo's for being interested in the cardinal numbers. However, there's a few errors in the question, so lets just try to set the record straight.

The (currently standard) collection of set-theoretic principles is called ZFC. So rather than asking "what is true of the cardinal numbers?" (vague question), let us ask "what can ZFC prove about the cardinal numbers?" (precise question).

ZFC proves the following sentences.

1. $\beth_0 = |\mathbb{N}|$
2. $\beth_1 = |\mathbb{R}|$ (but we cannot prove $\aleph_1 = |\mathbb{R}|$).
3. $\beth_0 < \beth_1$.
4. There exist infinitely-many infinite cardinal numbers.
5. There does not exist a set of all cardinal numbers.

To see why there ought to exist infinitely-many infinite cardinal numbers:

Step 1. Recall Cantor's theorem: $|X| < |\mathcal{P}(X)|$ for all sets $X$.

Step 2. Consider the following sequence. $$|\mathbb{N}|, |\mathcal{P}(\mathbb{N})|, |\mathcal{P}(\mathcal{P}(\mathbb{N}))|,...$$

(Of course, this is just the initial portion of the beth sequence: $\beth_0, \beth_1,\beth_2,\ldots$)

I might have a go answering some of your other questions a bit later.

• Thank you very much for this detailed answer. So from #5 I understand that there is no means by which we can count the number of infinite cardinalities, because we cannot define a set for them. Is that correct? – barak manos Mar 27 '14 at 12:35
• @barakmanos Ironically, even though the total collection is larger than any set, we can count them, in the generalized sense of ordinal "counting", which includes these magical "limit steps" which formalize the "$\dots$" in $$\aleph_0,\aleph_1,\aleph_2,\dots,\aleph_\omega, \aleph_{\omega+1},\dots,\aleph_{\omega\cdot2},\dots,\aleph_{\omega^2},\dots, \aleph_{\omega^\omega},\dots.$$ See the wiki article on Ordinal numbers for details. – Mario Carneiro Mar 27 '14 at 13:45
• Commentary on the sets $|\Bbb N|,|{\cal P}\Bbb N|,|{\cal P}^2\Bbb N|,\dots$: The set $\bigcup_{n\in\omega}{\cal P}^n\Bbb N$ is a set (being a countable union of sets), and is clearly larger than any set in the union. Thus it cannot be smaller than $\beth_n$ for any $n\in\omega$. In fact, this set has cardinality $\beth_\omega$. – Mario Carneiro Mar 27 '14 at 13:53
• @barakmanos, kind of. Ill edit with clarification when I can find the time. – goblin GONE Mar 27 '14 at 14:52
• Can ZFC actually prove that that those iterated power sets all exist and are all distinct? – Jack M Mar 27 '14 at 18:10

The "set" of cardinalities is not a set. It is a class. Every set is a class, but some classes are sets and some aren't. There are things that you can make with a set and not with a class, and one of these things is defining its cardinality. But a class that is not a set always has infinitely many elements.

So:

1. There is no "set of cardinalities". Thus, there is no cardinality of the set of cardinalities. I'm not sure, but this sounds like the typical thing that drives to paradox; perhaps the cardinality of the sets of cardinalities would be bigger than any other cardinality (that would be the reason that it doesn't exist).
2. There are infinitely many cardinalities. A simple proof has been written by JiK in his/her comment.

Look here for some details.

• OK, I wasn't aware of the term class. Thanks. – barak manos Mar 27 '14 at 11:23
• proper class. – user642796 Mar 27 '14 at 11:31
• Personally, I've come to believe that talk of "classes" just confuses things. Lets talk about models and theories instead. ZFC (a theory) proves there does not exist a set of all cardinal numbers. By the completeness theorem for first order logic, this is equivalent to the statement that the cardinal numbers of a universe cannot be internalized to that universe. More precisely, the statement is that "if $(M,\in')$ is a model of ZFC and $K \subseteq M$ is the set of $M$-cardinal numbers, then there is no $k \in M$ such that $K = \{m \in M \mid m \in' k\}.$" – goblin GONE Mar 27 '14 at 12:23

Q1.yes.there is not only an infinite number of infinite cardinalities,but the collection of infinite cardinals is a proper class,that is too big to be a set,bigger than all cardinals. Q2.it is surely uncountable,and also bigger than all uncountable cardinals. aleph-1=cardinality of R is true if and only if CH is true,otherwise R can have cardinality aleph-2,aleph10,aleph-1232337312,or other alephs.cardinals have opertion + and *,a+b=a*b=max{a,b} for all infinite cardinals.- and / can not be defined. you can have unimaginable big alephs,such as aleph-aleph-aleph-.......,aleph-0 times,that is the smallest cardinal such that aleph-a=a.BUT it is consisitent with ZFC,it is still smaller than the cardinality of R,2^aleph-null!!!aleph-x just means the xth infinite cardinal.