Pete's excellent notes have correctly explained that there is no set containing sets of unboundedly large size in the infinite cardinalities, because from any proposed such family, we can produce a set of strictly larger size than any in that family.
This observation by itself, however, doesn't actually prove that there are uncountably many infinities. For example, Pete's argument can be carried out in the classical Zermelo set theory (known as Z, or ZC, if you add the axiom of choice), but to prove that there are uncountably many infinities requires the axiom of Replacement. In particular, it is actually consistent with ZC that there are only countably many infinities, although this is not consistent with ZFC, and this fact was the historical reason for the switch from ZC to ZFC.
The way it happened was this. Zermelo had produced sets of size $\aleph_0$, $\aleph_1,\ldots,\aleph_n,\ldots$ for each natural number $n$, and wanted to say that therefore he had produced a set of size $\aleph_\omega=\text{sup}_n\aleph_n$. Fraenkel objected that none of the Zermelo axioms actually ensured that $\{\aleph_n\mid n\in\omega\}$ forms a set, and indeed, it is now known that in the least Zermelo universe, this class does not form a set, and there are in fact only countably many infinite cardinalities in that universe; they cannot be collected together there into a single set and thereby avoid contradicting Pete's observation. One can see something like this by considering the universe $V_{\omega+\omega}$, a rank initial segment of the von Neumann hierarchy, which satisfies all the Zermelo axioms but not ZFC, and in which no set has size $\beth_\omega$.
By adding the Replacement axiom, however, the Zermelo axioms are extended to the ZFC axioms, from which one can prove that $\{\aleph_n\mid n\in\omega\}$ does indeed form a set as we want, and everything works out great. In particular, in ZFC using the Replacement axiom in the form of transfinite recursion, there are huge uncountable sets of different infinite cardinalities.
The infinities $\aleph_\alpha$, for example, are defined by transfinite recursion:
- $\aleph_0$ is the first infinite cardinality, or $\omega$.
- $\aleph_{\alpha+1}$ is the next (well-ordered) cardinal after $\aleph_\alpha$. (This exists by Hartog's theorem.)
- $\aleph_\lambda$, for limit ordinals $\lambda$, is the supremum of the $\aleph_\beta$ for $\beta\lt\lambda$.
Now, for any ordinal $\beta$, the set $\{\aleph_\alpha\mid\alpha\lt\beta\}$ exists by the axiom of Replacement, and this is a set containing $\beta$ many infinite cardinals. In particular, for any cardinal $\beta$, including uncountable cardinals, there are at least $\beta$ many infinite cardinals, and indeed, strictly more.
The cardinal $\aleph_{\omega_1}$ is the smallest cardinal having uncountably many infinite cardinals below it.
Ordinals are not cardinals.
Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.
So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.
Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.
But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $\aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $\aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.
This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.
Okay, what's the point of all that?
The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".
When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.
So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $\omega$ is the appropriate notation, since it denotes an ordinal, rather than $\aleph_0$ which denotes a cardinal.
Between $\aleph_\omega$ and $\aleph_{\omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".
Note that $\aleph$ numbers are not defined by power sets, these are $\beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.
Best Answer
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.