# Are there cardinalities beyond Aleph 2? [duplicate]

I'm familiar with Aleph Null, Aleph 1, and Aleph 2 but are there greater [uncountable] infinities beyond (examples) Natural numbers, Real numbers, and the number of curves that can pass through a point? If so, what are some tangible examples of these cardinalities: Aleph 3, 4, ...

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• In fact, maybe the cardinality of real numbers is already more than $\aleph_2$. But maybe it is just $\aleph_1$. – Hagen von Eitzen May 24 '18 at 18:06
• Ugh. I am starting to really hate that frigging book. – Asaf Karagila May 24 '18 at 18:08
• @AsafKaragila Just relax. Count to three... – David C. Ullrich May 24 '18 at 18:29
• @AsafKaragila Which book do you mean? – Cornman May 24 '18 at 18:49
• @Cornman: The one by Gamow. – Asaf Karagila May 24 '18 at 18:50

The next one after $\aleph_2$ is $\aleph_3$. There's also $c$, the cardinality of the reals. In fact $c=\aleph_\alpha$ for some $\alpha$, but nobody knows which one...

Regading your new request for "tangible examples": In fact $\aleph_0$ is the cardinality of the integers, which I gather is what you mean by a tangible example. But no, $\aleph_1$ is not the carinality of the reals.

So what are these aleph things? One can prove that given a cardinal there is a smallest larger cardinal. So $\aleph_1$ is by definition the smmallest uncountale cardinal, $\aleph_2$ is by definition the smallest cardinal larger than $\aleph_1$, etc. That's as "tangible" as it gets.

Finally, regarding "Natural numbers, Real numbers, and the number of curves that can pass through a point?": Although you didn't actually say so, this sounds like you think the cardinality of that set of curves is larger then the reals. This is not so. If a "curve" is the graph of a continuous function, then the cardinality of the set of all curves, passing through a given point or not, is just $c$.

• I thought c and ℵ1 were the same because it is the cardinality of reals. I don't think the continuum hypothesis requires the axiom of choice either. – poetasis May 24 '18 at 18:08
• @poetasis No. That statement is known as the continuum hypothesis, and is known to not be decidable from the usual axioms of set theory. Indeed, in a precise sense ZFC can't prove anything about the exact value of the continuum in terms of the $\aleph$-hierarchy except that it can't have countable cofinality. – Noah Schweber May 24 '18 at 18:08
• @poetasis No. I've heard that there are well-known popular books that state that $c=\aleph_1$, but it's not so. At least not provably so; it's been proved that you can't prove $c=\aleph_1$ and you also can't prove $c\ne\aleph_1$. – David C. Ullrich May 24 '18 at 18:10
• @poetasis "don't think CH requires AC" sounds like you're saying AC is not needed to prove CH. In fact you can't prove CH with or without AC. – David C. Ullrich May 24 '18 at 18:20
• It is slightly more tangible to say that $\aleph_1$ is the cardinality of the set of isomorphism classes of total orders on $\mathbb N$ that are well-orders. (But still nothing that really engages the imagination). – Henning Makholm May 24 '18 at 18:38

Yes. Just keep taking power sets of power sets ad infinitum. There is no largest cardinal.

To conceptualize larger cardinals, you can use power sets, for sure. If you want something larger than the cardinality of curves passing through a point, consider subsets of the set curves passing through a point.

$$|\mathcal{P}(A)|>|A|$$

Another way of looking at this approach is to consider the set of functions from a set of one cardinality to, for example, the set $\{0,1\}$. This set is equivalent to the power set of your original set, and its cardinality is greater. In symbols: $$|\left\{f|f:A\to\{0,1\}\right\}|>|A|$$