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.
You could say that $\frac{1}{\infty} = 0$, so $1-\frac{1}{\infty} = 1$. But then, you're stretching the definition of division past breaking point - division as you know it isn't defined for infinity, so the answer is undefined. Otherwise, you can quickly get yourself into a pickle and end up saying 1=2.
Arithmetic operators - add, subtract, divide, multiply, raise to the power of - are defined on a particular set of numbers: such as real numbers, or complex numbers.
The set you use for definition, will determine what you can and can't say meaningfully. Typically (but not always), infinity is excluded from that set.
If we take the set of real numbers, and look at "raise to the power of", then $1^x$ is equal to 1 for any x, as x -> infinity. So in that case, you could have a convention of saying that $1^\infty = 1$. But $\frac{1}{1} = 1$, so $1^{-\infty}$ would also equal 1. However, when you go about defining these new conventions, you have to be extremely careful - sometimes, a convention will seem obvious, but if you run with it, you end up seeming to prove 1=2, which means that your convention wasn't that helpful.
Let's compare with raising to the power 0.5, i.e. taking the square root. $-1^{0.5}$ is undefined when we are working on the reals - so, just as dividing by infinity, you can't include it in your arithmetic. Only when you expand to the complex numbers and extend your definition of the arithmetic operators to cope, can you say something meaningful about $(-1)^{0.5}$
Similarly, the reals and the complex numbers each exclude infinity, so arithmetic isn't defined for it.
You can extend those sets to include infinity - but then you have to extend the definition of the arithmetic operators, to cope with that extended set. And then, you need to start thinking about arithmetic differently. If you want to learn more about that, then there are lots of friendly places on the web to get into the work of Cantor on the different types of infinity (of which there are an infinite number of different infinities).
Best Answer
Often traditional reasoning from arithmetic breaks down when you try to think about infinity, and percentages are no exception.
Examples: You agree to pay me $B(n)/n$ dollars per year, where $B(n)$ is a function giving your total wealth after $n$ years (thanks!). So as time goes by, the portion of your income you're paying me is $1/n$ (or $100/n$ percent). As the years pass, this dwindles down to an "infinitely small percentage."
As time goes by, let's say you're a hard worker and immortal, and you earn more and more, say $B(n) = n$. Then every year you're paying me a dollar. The PERCENTAGE of your income that you are paying me is getting smaller and smaller each year, over time, but you're giving me a dollar a year. So in this case "an infinitely small percentage of infinity" is one.
In experiment two, you're still immortal, but you don't work as hard. Now $B(n) = \sqrt{n}$. Then every year you're paying me $\frac{1}{\sqrt{n}}$ dollars. As time goes by, your income is still going to infinity, but what you're paying me is going to dwindle down to zero. So "an infinitely small percentage of infinity" is zero.
In our final experiment, you're immortal and work really hard, and $B(n) = n^2$. Now you're paying me $n$ dollars a year, and an infinitely small percentage of infinity is infinity.