>>15
This is a very hand-wavey explanation. A much more straight-forward one:
Let X be a set, and P(X) be the set of all subsets of X. Suppose there exists a one-to-one mapping f from X to P(X). Let S = {x in X such that x is not in f(x)}. S is an element of P(X), so there must be an element y in X such that f(y) = S. We know that either y is in S or y is not in S. If y is in S, then y is in f(y), which means y can not be in S. If y is not in S, then y is in f(y), which means y must be in S. This is a contradiction, and since the only assumption we have made is that the one-to-one mapping f exists, we conclude that it does not exist.