There are two ways of comparing the number of elements in a set. You either count them, and see which one is bigger, which is fine for finite sets, or you can try and set up a one to one correspondence between the elements.

Kolmogrov gives the example of comparing the number of students is a classroom to the number of chairs: if there are still chairs left over when eveyone is sat down, then there are more chairs than students, and if there are still students standing up when everyone is sat down, then there are more students than chairs. And, in fact, this works well for infinite sets.

The fact that \(\displaystyle |A| < |\mathcal{P}(A)|\) means that we can match every element of \(\displaystyle A\) to an element of \(\displaystyle \mathcal{P}(A)\) like students to chairs, with some elements in \(\displaystyle \mathcal{P}(A)\) to spare, which intuitively means that the set \(\displaystyle \mathcal{P}(A)\) is 'bigger' than \(\displaystyle A\). The fact that there is no biggest set means that there is no set which cannot be 'fit inside' another set in this way: no matter how many students we have, we can always find a classroom to sit them all __with some chairs left over__.

In particular, suppose that \(\displaystyle A\) is the biggest possible set, then \(\displaystyle \mathcal{P}(A)\) is bigger, a contradiction. To answer the question, \(\displaystyle |\mathcal{P}(\mathbb{R})|>|\mathbb{R}|\), so yes there is such an \(\displaystyle S\).

And as for your second question, the fact that \(\displaystyle (0,1)\) and \(\displaystyle \mathbb{R}\) are numerically equivalent doesn't mean that any infinite subset of \(\displaystyle \mathbb{R}\) is numerically equivalent to \(\displaystyle \mathbb{R}\) because \(\displaystyle \mathbb{N}\subset\mathbb{R}\) yet \(\displaystyle \mathbb{R}\) is not countable. The *purpose *of the theorem is anybody's guess, you might as well ask what the meaning of life is!