We can make a one-to-one correspondence between \(\mathscr{S}\) and the set of singleton subsets of \(\mathscr{S}\); each element \(s\) of \(\mathscr{S}\) gets assigned to the singleton \(\{s\}\). This is a valid one-to-one correspondence, between \(\mathscr{S}\) and a subset of \(P(\mathscr{S})\). So we know that \(\mathscr{S}\) has to be no larger than \(P(\mathscr{S})\).
Next we will show that \(P(\mathscr{S})\) is not the same size as \(\mathscr{S}\), and this will be enough to conclude that \(| P(\mathscr{S})| > |\mathscr{S}|\).
To show \(\mathscr{S}\) and \(P(\mathscr{S})\) are not the same size, we must show that there can be no one-to-one correspondence between them. Here it's helpful to think of assignments between sets as functions.
A function \(f\) from \(\mathscr{S}\) to \(P(\mathscr{S})\) is an assignment rule. For each element \(x\) in \(\mathscr{S}\), it assigns an element \(f(x)\) in \(P(\mathscr{S})\). As far as we're concerned here, that is all a function is -- an assignment rule. But in this case, since \(P(\mathscr{S})\) is the set of subsets of \(\mathscr{S}\), a function from \(\mathscr{S}\) to \(P(\mathscr{S})\) seems a little bizarre. To each element \(x\) in \(\mathscr{S}\), the function \(f\) assigns a subset \(f(x)\) of \(\mathscr{S}\).
A one-to-one correspondence is a special type of function; it's a function that makes sure everything gets hit once and only once. In other words, a function \(f\) from \(\mathscr{S}\) to \(P(\mathscr{S})\) is a one-to-one correspondence if every element in \(P(\mathscr{S})\) gets hit exactly once by \(f\). More precisely, \(f\) is a one-to-one correspondence if, for every subset \(Z\) of \(\mathscr{S}\) -- i.e. every element \(Z\) of \(P(\mathscr{S})\) -- there is one and exactly one element \(t\) in \(\mathscr{S}\), so that \(f(t) = Z\).
What we will show is that any function \(f\) from \(\mathscr{S}\) to \(P(\mathscr{S})\) fails to hit everything. In other words, given a function \(f\), we'll show that we can construct a special subset \(Y\) of \(\mathscr{S}\) in such a way that we are sure no element of \(\mathscr{S}\) gets assigned to \(Y\) by \(f\).
So now suppose we have a function \(f\) from \(\mathscr{S}\) to \(P(\mathscr{S})\). Consider the following subset of \(\mathscr{S}\), i.e. element of \(P(\mathscr{S})\).
\(Y= \{\) elements \(s\) in \(\mathscr{S}\), such that \(s\) is not an element of \(f(s) \}\)
This may seem like an odd way to define a subset, describing it in words. But it is perfectly acceptable (at least in this basic version of set theory we are using). The words clearly give a condition for deciding what elements of \(\mathscr{S}\) should be included in \(Y\). Mathematicians do this all the time. For example, we could describe the set of even natural numbers as
\(2\mathbb{N} = \{\) all natural numbers that are divisible by two\( \}\).
What did we just conclude? For each element \(s\) in \(\mathscr{S}\), the sets \(f(s)\) and \(Y\) are not the same. Now, thinking of \(f\) as a function from \(\mathscr{S}\) to \(P(\mathscr{S})\), this says that there is no element of \(\mathscr{S}\) getting assigned to \(Y\)! The set \(Y\) is the special element of \(P(\mathscr{S})\) that is not -- cannot -- be hit by the function \(f\).
This trick works for any function from \(\mathscr{S}\) to \(P(\mathscr{S})\). We can always construct a special element of \(P(\mathscr{S})\) that does not get hit by the function. The conclusion is that any function from \(\mathscr{S}\) to \(P(\mathscr{S})\) will fail to be a one-to-one correspondence; there can be no one-to-one correspondence between \(\mathscr{S}\) and \(P(\mathscr{S})\); and so \(\mathscr{S}\) is strictly smaller than \(P(\mathscr{S})\).