Because mathematicians like to be pedantic, we also consider \(\{a, b, c\}\) to be a subset of \(\{a, b, c\}\). In other words, every set is a subset of itself. There's not much deep to this statement; it's just how we define the term subset. It's also handy to allow for the empty subset, the set that has nothing in it. We'll write this \(\{ \}\). This is also an acceptable subset of \(\mathscr{S}\).

Adding the full and empty subsets, this means that \(\{a, b, c\}\) has a total of 8 subsets.

Thus the power set \(P(\{a, b, c\})\) in this example is this set:

\[\{ \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{a,b\}, \{ \}, \{a,b,c\} \}.\]

**But we want to talk about infinite sets.** Suppose now that \(\mathscr{S}\) is some infinite set. We can build \(P(\mathscr{S})\), its power set, whose elements are all the subsets of \(\mathscr{S}\). I'm claiming that \(| P(\mathscr{S})| > | \mathscr{S}|\).

Everything in \(\mathscr{S}\) appears in \(P(\mathscr{S})\), as a singleton set. In our finite example, \(a\) and \(b\) and \(c\) in \(\mathscr{S}\) appeared as \(\{a\}\) and \(\{b\}\) and \(\{c\}\) in \(P(\mathscr{S})\). So it seems intuitively clear that \(| P(\mathscr{S})|\) should be at least as big as \(|\mathscr{S}|\). But \(P(\mathscr{S})\) has a lot of other subsets, besides the singleton sets.