Oh no, it's him again!! Well tough tits, dear friends, but I promise you will enjoy this, however before the fun starts, we have to do a bit of homework.
The
cardinality of a set is simply its number of elements.
If

is a set, then its powerset

is simply the set formed from
all its subsets. Thus, bearing in mind that

are always subsets of

(

is the empty set, btw) then, when

, then

.
Note that the elements in the powerset are subsets of the set - this will be important.
There is a Theorem of Cantor that says that the cardinality of

is always strictly greater than the cardinality of

.
In the example above, this is not hard to see; the cardinality of

is 3, that of

is

. In fact, using all available fingers and toes, and those of our wives and girlfriends (should they be different), it is not hard to see that, in general, for any set

of finite cardinality

, then the powerset

has cardinality
But what if the cardinality of our set is infinite? What do we make of the assertion, say, that

? Surely this is madness?
One last thing, of MAJOR importance (sorry for shouting).
A set is said to be
countable iff it is isomorphic to a subset of the natural (i.e. "counting") numbers

. Since, from the above,

is always a subset of itself, and is infinite, we have the brain-curdling expression "countably infinite". (That's yet another reason to love mathmen).
Let's assume our set

is countably infinite in this sense.
Let's also assume that

is countably infinite in the same sense, that is, in accord with intuition,

. We will find a contradiction
So, since

is countable, we can index each and every element by an element of

(this is due to our isomorphism). Call the

-th element

. (Note that the choice of ordering is arbitrary, but, having chosen, we had jolly well better stick with it)
Assuming that

is also countably infinite, then, to each element here we can also assign an index. So let's write a list of these elements (subsets of

, remember) and call the

-th member of this list as

.
Now form the set

by the rule that

iff
Now

is obviously a subset of

, an element in

, so is eligible to be in our list. Let's call this list element as

, so by our rule we have that

iff

.
But hey!

so we arrive at the breath-taking conclusion:

iff

.
Wow! Let's write a paper on that. But wait....
For every subset in

(that is every element in

) I can find its complement, that is, there is the element in

comprising those elements in

that are
not in

Let's call it as

.
Under the assumption that our powerset is countable, I can add

to our list, call it the

-th member, and apply the same rule as before:

iff

.
But

, so we conclude that

iff

. This is clearly nuts, so the assumption that the powerset is countable must be false. Cantor's Theorem is thereby proved.
Way too long, but really sweet, wouldn't you say?