Much like I wrote in Cardinal number subtraction, if $\kappa$ and $\lambda$ are two $\aleph$-numbers, it might be possible to define division, but this definition would have to be limited and awkward.
If $\kappa$ and $\lambda$ are both regular cardinals and $\kappa<\lambda$ then every partition of $\lambda$ into $\kappa$ many parts would have to have at least one part would be of size $\lambda$. In a sense this means that $\frac\lambda\kappa=\lambda$. This is indeed the case with $\aleph_1/\aleph_0$, both are regular cardinals are $\aleph_0<\aleph_1$.
If, however, $\kappa=\lambda$ this is no longer defined, since $\kappa=2\cdot\kappa=\aleph_0\cdot\kappa=\ldots=\kappa\cdot\kappa=\ldots$, so there can be many partitions of $\lambda$ into $\kappa$ many parts, and in each the parts would vary in size (singletons; pairs; countably infinite sets; etc.)
When $\lambda$ is a singular limit cardinal, e.g. $\aleph_\omega$ this breaks down completely, since singular cardinals can be partitioned into a "few" "small" parts. In the $\aleph_\omega$ case these would be parts of size $\aleph_n$ for every $n$, which make a countable partition in which all parts are smaller than $\aleph_\omega$.
The only reasonable way I can think that cardinal division can be defined would have to consider the Surreal numbers, and the embedding of the ordinals in them. However this will not be compatible with cardinal arithmetic at all (the surreal numbers form a field).
I should also remark that your reasoning for $\aleph_1/\aleph_0$ being $\aleph_1$ is invalid. First note that neither is a real number, and that it is possible that $\aleph_1$ is much smaller than the cardinality of the real numbers (so between two natural numbers there are a lot more real numbers). Secondly, note that between two rational numbers there are also infinitely many rational numbers - does that mean $\aleph_0/\aleph_0=\aleph_0$?
However your rationale is not that far off, as I remarked in the top part of the post, if you take a set of size $\aleph_1$ and partition it into $\aleph_0$ many parts you are guaranteed that at least one of the parts would have size $\aleph_1$.
Further reading:
- Cofinality of cardinals
- Cofinality and its Consequences
- How to understand the regular cardinal?
- How far do known ordinal notations span? (Cantor normal form)
- Surreal and ordinal numbers
To understand the $\aleph$ numbers properly you need to understand the ordinal numbers first, at least a little bit.
The idea of an ordinal is to model the notion of a length of a queue to the bathroom in a party. There is an empty queue, then there is one person waiting, then another, and so on. But here we can talk about infinite queues as well. At some point we have an infinite queue, but everyone on that queue is just a finite number of people away from the bathroom. This is the same as the natural numbers, there are infinitely many of them, but each is preceded by just a finite number.
Then comes the poorest schmoe you had the luck to meet. And he has to wait for all those infinitely many people to use the bathroom before he can. Poor guy. But it could be worse, someone comes and takes place after this poor schmoe. And then another and another and soon enough we have two copies of the natural numbers stacked one on top of another.
And we can continue in this fashion. At some point it becomes impossible to properly visualize. It's just a very very long long to the bathroom, and you really take pity on whoever is coming to stand in this queue next.
But after you've done this for every countable queue possible, you have an uncountable queue, and each person in that queue is just a countable number of people away. But here comes a new person, and she has to wait uncountably many people before she can see the inside of that bathroom. That girl is standing on the first uncountable point of the queue. And sure enough, we can continue again, and extend and enlarge and so on and so forth.
Okay, so how do we get from this to the $\aleph$ numbers? The $\aleph$ numbers tell us how many people are in the queue, not how long it is. For finite queues the two notions coincide, but for infinite queues they do not. It is clear that the queue described by $\Bbb N$, and the queue described by adding one more after that are both countable. If that new person had cut in line, he would piss a lot of folks, but the length of the queue remains the same.
So $\aleph$ numbers come to tell us how many people are standing in that line. But in order to make this a well-defined notion, we use the shortest queue that can hold that many people. So $\aleph_0$ is the size of the shortest queue that is infinite, namely the queue where each person has only finitely many predecessors; $\aleph_1$ is the size of the queue where there are uncountably many people, but each has at most $\aleph_0$ predecessors; $\aleph_2$ is the size of the shortest queue where each person has at most $\aleph_1$ predecessors.
This goes on, and we can easily see how a queue holding more than $\aleph_1$ and more than $\aleph_2$ and more than $\aleph_n$, for every finite $n$, is formed. But we can continue, and the next cardinal is exactly that of a line where each person has at most $\aleph_n$, for some $n$, predecessors. This is a limit cardinal and it is denoted by $\aleph_\omega$, since $\omega$ is the first infinite ordinal (it denotes the natural numbers).
And so we can continue, and show that if $\alpha$ is an ordinal, then there is a cardinal with exactly $\alpha$ cardinals preceding it, and we call this cardinal $\aleph_\alpha$.
So how many are there? Well, this mapping is a bijection between the ordinals and the $\aleph$ numbers. Since the collection of all ordinals is too large to even be considered a set, the collection of all cardinals is too large to be a set as well. And we call these collections "proper classes" to say that they are not sets, but collections which while large enough, are still collections we can talk about.
Best Answer
If by $\{ \aleph_0, \aleph_1, \aleph_2, \dots \}$ you mean $\{ \aleph_n : n \in \mathbb{N} \}$, then this set is countable because it's indexed by the natural numbers. However, this set doesn't contain $\aleph_{\omega}$ or any larger alephs.
If by $\{ \aleph_0, \aleph_1, \aleph_2, \dots \}$ you mean 'the set[sic] of all (well-orderable) cardinals', then it's certainly not countable; in fact, it's too big to be a set! (Hence the 'sic'.) There is an aleph for each ordinal $\alpha$, and there are class-many ordinals.
It should also be noted that, unless you accept the generalised continuum hypothesis, it is not necessarily the case that $\aleph_{\alpha} = |\mathscr{P}^{\alpha}(\mathbb{N})|$ for all $\alpha$, as you suggest in your question.