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.
The notion of "countably infinite" is well named. Another word you can use is "enumerable," which is even more descriptive in my opinion.
I understand your intuition on the subject and I see where you're coming from. Let me try to give you some insight into the agreements about infinity that have been reached over the years by the mathematicians who've worked on this problem. (This is what I wish someone had explained to me.)
"Countably infinite" just means that you can define a sequence (an order) in which the elements can be listed. (Such that every element is listed exactly once.)
The natural numbers are the most naturally "countable"—they're even called "counting numbers"—because they are the most basic, natural sequence. The word "sequence" itself is wrapped up in what we mean by natural numbers, which is just one thing following another, and the next one coming after that, and the next one after that, and so on in sequence.
But the integers are countable as well. In other words, they are enumerable. You can specify an order which lists every integer exactly once and doesn't miss any of them. (Actually it doesn't matter, for the meaning of "countable," if a particular element gets listed more than once, because you could always just skip it any time it appears after the first time.)
Here is an example of how you can enumerate (count, list out) every single integer without missing any:
$0, 1, -1, 2, -2, 3, -3...$
The rational numbers are also countable, again because you can define a sequence which lists every rational number. The fact that they are listed means (at the same time) that they are listed in a sequence, which means that you can assign a counting number to each one.
The real numbers are not countable. This is because it is impossible to define a list or method or sequence that will list every single real number. It's not just difficult; it's actually impossible. See "Cantor's diagonal argument."
This will hopefully give you a solid starting point to understanding anything else about infinite sets which you care to examine. :)
Best Answer
Clearly the group $\Bbb Z$ is generated by $\langle 1\rangle$, but also by $\langle\Bbb Z\rangle$. So from your point of view, an infinite set seems to be the same size as a singleton.
Set theory is the framework for comparing infinite sets, and in the standard context of $\sf ZFC$, uncountable implies, and in fact synonymous with, "larger than countable".
You are forgetting that the process of "spanning" here is such that the cardinality can grow larger, also because it involves taking infinite sums.